Problem with multiple comparisons. Can someone explain how to deal with this in the situation described?

Question

Charles Yang Zheng · Answer

In most natural occurring situations, it will be better to use multiple comparisons on all attributes than to select a subset of attributes completely at random.  At the very least, you will usually not do much worse by testing all hypotheses.

An example in which it would be better to randomly select attributes is if, for some reason, the non-null attributes only have "small" p-values but never have "extremely small" p-values.  But even this strange possibility can be ruled out if, for example, your data points have a non-negligible amount of measurement error.

On the other hand, if you do have prior knowledge about which attributes are likely to be important, *then* you have an important choice to make.  In many cases you will be better off only testing those pre-selected attributes.

It is easy to get confused if you view frequentist statistical procedures in terms of "truth" or "evidence."  For example, rolling a twenty-sided die provides a universal hypothesis test at p=0.05, but you should not interpret rolling a 1 as "evidence" that the sun has exploded (relevant xkcd: Frequentists vs. Bayesians [ http://xkcd.com/1132/ ]).  Instead, you should view frequentist statistical procedures as completely specified algorithms and then judge them by their average performance on random instances of the data.

As a result of this, the interpretation of the same data may vary depending on which algorithm you choose.  (This is indeed counterintuitive, and it is one justification for using a more intuitive Bayesian approach.)

Here are three different reasonable procedures you could have used for your thought experiment--all potentially producing differing interpretations of your data.

Procedure 1:  Choose a random attribute and do a classical hypothesis test at level [math]\alpha[/math].  Then stop.

Procedure 2:  Choose a random attribute and do a classical hypothesis test at level [math]\alpha[/math].  If it is significant, choose another random attribute and test it at level [math]\alpha[/math].  Continue in this way until you accept the null hypothesis.

Procedure 3:  Test all the hypothesis, adjusting for multiple comparisons.

Trevor Brothers · Answer

I think using permutation statistics can help you get an intuitive handle on multiple comparison problems like this.

Let's say that, hypothetically, brain scans are a poor tool for diagnosing our three clinical groups (A, B, and C), and therefore, none of our data points are going to help us predict someone's diagnosis. If this were true, then any differences we observed in our current data set arose just by chance. In fact, we would expect to see differences of the same size if we assigned our brain data to the three groups completely at random.

Using permutation statistics we can test this assumption empirically. First, assign each subject to a random group. Then run all your same tests again and look for the largest effect size (t-value) that you get for any measure in any brain region. Given your large number of variables and brain regions this t-value is likely going to be quite big - (and that just occurred by chance!)

So the real question is, is your original data so outstanding that it convinces you that these subjects weren't assigned to the groups at random? Are the results so shockingly different from all the possible random assignments (permutations) that we have to reject the null hypothesis?

Again let's test this empirically. Shuffle your original subjects again, and again record the largest t-value. Now do it 1000 more times (automated of course using a free program like R) getting a distribution of your maximum "by chance" t-values, which you can then plot like a histogram.

Now comes the fun part - compare this histogram to the maximum t-values in your real dataset. How do they stack up? If all your real t-values are sitting right in the middle of the distribution then there's nothing convincing in this dataset - either the technique doesn't work, or you don't have enough data to answer such a complex question (do any of these three groups differ on any of these measures in any of these brain regions). BUT, if any of your real t-values are waaaay off in the tails (a 5% criteria works just like your regular alpha level) then you have convincing evidence that these differences are real. In fact, any tests beyond the 5% cutoff can be safely regarded as significant.

This technique might not give you the answer you're looking for, but it will give you an appreciation for what ridiculously (t=6.5!?) significant looking results you can see sometimes when performing extensive multiple comparisons. It's always hard finding a needle in a haystack, and this might give you an appreciation for why.

Anonymous · Answer

For a brief moment, I want you to sit down and think - who you are and who you are becoming.

( This is what I tell myself, when I tend to compare myself to others).

Life needs to be lived intentionally and thoughtfully. Not as a reaction to what is happening around us.

I would like to share an experience.

I have recently Joined Martial arts ( karate). My Class has all ages, and Students with different levels of belts.

Learning the White Belt kata wasn’t hard. I mastered it real fast. But when I was doing with the rest of the White Belt students, Although I was doing the right move and I know my kata well, I was trying to compare myself with others and follow them and I was making blunders. Because my focus was on their Pace. And I was loosing my stances and my Moves.

I Had to Stop. And I had to tell myself to stop looking at them.

Once I Started focussing on my own feet, My own stances, and my brain didn’t get confused anymore. I started working on myself instead of wasting my time trying to watch others.

I had to feel the Power within myself. I took back my power, control of my focus. So, the Story needs to end right.? I Did Really well at my grading and now I’m a confident Green Belt holder.

Grades, Clothes, Cars, income, Job title, house, Upvotes, likes, followers - the number of categories in which we compare ourselves with others is infinite. Comparing yourself to others and what they have causes misery. It triggers self- doubt, frustration and in-adequacy.

You do not have the complete picture of what others lives are about. There is no point in comparing yourself with others. You wouldn’t achieve anything by watching your neighbours grass. Nurture your own grass. Your personalities, your purpose, your struggles, your story is so different to others, Then Why compare?

So, What kind of Comparison could be healthy for you?

For example there are some really amazing people who Spread kindness by doing various volunteering work. They make a difference in people’s lives and I want to be like them!

Turn your comparison to inspiration. Be inspired by other people’s stories of how they got their and about their hard work.

Don’t worry about your neighbours Tesla or Your colleagues Job promotion. You will end up with stomach Ulcers and Doctor’s Bills and nothing else. Do not waste time in comparing with others. Accept who you are. Embrace your story.

Compete with yourself.

What matters is your mindset, attitude, and where you’re going. Don’t waste your Power watching other People. Take back your power. Invest in your growth. Become a better person than yesterday.

Personality begins where comparison leaves off. Be unique. Be memorable. Be confident. Be proud.” ~Shannon L. Alder ♥️

Anonymous · Answer

Sixteen data points on three groups, pre and post experiment? You don't have enough data for anything. I would get the theory straight in your head and simplify it to the key elements, then control as much of the extraneous variables as possible in an experimental setting, and run the subjects one at a time until you have sufficient data. Use Bayesian stats if you can manage it, but to be honest if you have to look at the stats to find a significant difference, it isn't there.

Dushka Zapata · Answer

In yoga I used to frequently look at what other people were doing and feel frustrated and inadequate. Others could always do things I couldn’t do.

When I stopped looking at other people I could focus more on myself, on my own breathing, on where I was at on that day, my own progress, on the subtle balance of effort and ease.

I felt peace.

To stop comparing my life to others I don’t only stop looking to what others do - I focus on myself, work on myself.

I have so much work to do. So much.

Any time or energy I spend looking at others is time I don’t spend on my own development.

Mark Gritter · Answer

Consider the humble binomial coefficient [math]{ n \choose k }[/math], the number of ways to choose [math]k[/math] distinct items from a set of [math]n[/math] elements.

What is the number of ways to choose three pizza toppings from the eight listed on a menu? [math]{8 \choose 3} = 56[/math]. (Or, if you want to be difficult, there are other possible answers. [ https://www.quora.com/In-how-many-ways-can-8-toppings-be-selected-to-make-a-3-topping-pizza ])

[code]%3E%3E%3E pprint.pprint( list( itertools.combinations( ["pepperoni", "sausage", "bacon", "chicken", "mushroom", "black olive", "green pepper", "pineapple" ], 3 ) ) )
[('pepperoni', 'sausage', 'bacon'),
 ('pepperoni', 'sausage', 'chicken'),
 ('pepperoni', 'sausage', 'mushroom'),
 ('pepperoni', 'sausage', 'black olive'),
 ('pepperoni', 'sausage', 'green pepper'),
 ('pepperoni', 'sausage', 'pineapple'),
 ('pepperoni', 'bacon', 'chicken'),
 ('pepperoni', 'bacon', 'mushroom'),
 ('pepperoni', 'bacon', 'black olive'),
 ('pepperoni', 'bacon', 'green pepper'),
 ('pepperoni', 'bacon', 'pineapple'),
 ('pepperoni', 'chicken', 'mushroom'),
 ('pepperoni', 'chicken', 'black olive'),
 ('pepperoni', 'chicken', 'green pepper'),
 ('pepperoni', 'chicken', 'pineapple'),
 ('pepperoni', 'mushroom', 'black olive'),
 ('pepperoni', 'mushroom', 'green pepper'),
 ('pepperoni', 'mushroom', 'pineapple'),
 ('pepperoni', 'black olive', 'green pepper'),
 ('pepperoni', 'black olive', 'pineapple'),
 ('pepperoni', 'green pepper', 'pineapple'),
 ('sausage', 'bacon', 'chicken'),
 ('sausage', 'bacon', 'mushroom'),
 ('sausage', 'bacon', 'black olive'),
 ('sausage', 'bacon', 'green pepper'),
 ('sausage', 'bacon', 'pineapple'),
 ('sausage', 'chicken', 'mushroom'),
 ('sausage', 'chicken', 'black olive'),
 ('sausage', 'chicken', 'green pepper'),
 ('sausage', 'chicken', 'pineapple'),
 ('sausage', 'mushroom', 'black olive'),
 ('sausage', 'mushroom', 'green pepper'),
 ('sausage', 'mushroom', 'pineapple'),
 ('sausage', 'black olive', 'green pepper'),
 ('sausage', 'black olive', 'pineapple'),
 ('sausage', 'green pepper', 'pineapple'),
 ('bacon', 'chicken', 'mushroom'),
 ('bacon', 'chicken', 'black olive'),
 ('bacon', 'chicken', 'green pepper'),
 ('bacon', 'chicken', 'pineapple'),
 ('bacon', 'mushroom', 'black olive'),
 ('bacon', 'mushroom', 'green pepper'),
 ('bacon', 'mushroom', 'pineapple'),
 ('bacon', 'black olive', 'green pepper'),
 ('bacon', 'black olive', 'pineapple'),
 ('bacon', 'green pepper', 'pineapple'),
 ('chicken', 'mushroom', 'black olive'),
 ('chicken', 'mushroom', 'green pepper'),
 ('chicken', 'mushroom', 'pineapple'),
 ('chicken', 'black olive', 'green pepper'),
 ('chicken', 'black olive', 'pineapple'),
 ('chicken', 'green pepper', 'pineapple'),
 ('mushroom', 'black olive', 'green pepper'),
 ('mushroom', 'black olive', 'pineapple'),
 ('mushroom', 'green pepper', 'pineapple'),
 ('black olive', 'green pepper', 'pineapple')]
[/code]What is the coefficient on the [math]x^3[/math] term of [math](x+1)^8[/math]? It’s [math]{8 \choose 3} = 56[/math].

[math]x^8 + 8 x^7 + 28 x^6 + 56 x^5 + 70 x^4 + 56 x^3 + 28 x^2 + 8 x + 1[/math]

How many different ways can the positive integers [math]a, b, c, d[/math] sum to nine? The answer is again [math]{8 \choose 3} = 56[/math].

[code]%3E%3E%3E pprint.pprint( [ (a,b,c,d) for a in xrange( 1, 10 ) for b in xrange( 1, 10 ) for c in xrange( 1,10) for d in xrange( 1, 10 ) if a + b + c +d == 9 ] )
[(1, 1, 1, 6),
 (1, 1, 2, 5),
 (1, 1, 3, 4),
 (1, 1, 4, 3),
 (1, 1, 5, 2),
 (1, 1, 6, 1),
 (1, 2, 1, 5),
 (1, 2, 2, 4),
 (1, 2, 3, 3),
 (1, 2, 4, 2),
 (1, 2, 5, 1),
 (1, 3, 1, 4),
 (1, 3, 2, 3),
 (1, 3, 3, 2),
 (1, 3, 4, 1),
 (1, 4, 1, 3),
 (1, 4, 2, 2),
 (1, 4, 3, 1),
 (1, 5, 1, 2),
 (1, 5, 2, 1),
 (1, 6, 1, 1),
 (2, 1, 1, 5),
 (2, 1, 2, 4),
 (2, 1, 3, 3),
 (2, 1, 4, 2),
 (2, 1, 5, 1),
 (2, 2, 1, 4),
 (2, 2, 2, 3),
 (2, 2, 3, 2),
 (2, 2, 4, 1),
 (2, 3, 1, 3),
 (2, 3, 2, 2),
 (2, 3, 3, 1),
 (2, 4, 1, 2),
 (2, 4, 2, 1),
 (2, 5, 1, 1),
 (3, 1, 1, 4),
 (3, 1, 2, 3),
 (3, 1, 3, 2),
 (3, 1, 4, 1),
 (3, 2, 1, 3),
 (3, 2, 2, 2),
 (3, 2, 3, 1),
 (3, 3, 1, 2),
 (3, 3, 2, 1),
 (3, 4, 1, 1),
 (4, 1, 1, 3),
 (4, 1, 2, 2),
 (4, 1, 3, 1),
 (4, 2, 1, 2),
 (4, 2, 2, 1),
 (4, 3, 1, 1),
 (5, 1, 1, 2),
 (5, 1, 2, 1),
 (5, 2, 1, 1),
 (6, 1, 1, 1)]
[/code]What is the value of [math]T(8,3)[/math] for [math]T(x,y) = T(x-1,y-1) + T(x-1,y)[/math], with [math]T(0,y) = 1[/math] and [math]T(x,x) = 1[/math]? Probably no surprise at this point that it’s [math]{8 \choose 3} = 56[/math].

[code]%3E%3E%3E def t( x, y ):
	if y == 0:
		return 1
	if x == y:
		return 1
	return t( x-1, y-1 ) + t( x-1, y )

%3E%3E%3E t( 8, 3 )
56
[/code]Suppose you start at [math](0,0)[/math] and need to make your way to [math](5,3)[/math] by moving one step either north (up) or east (right) at a time. How many different paths can you take? It’s [math]{ 5+ 3 \choose 3 } = {8 \choose 3}.[/math] (Or your can count the other direction and get [math]{3  + 5 \choose 5}[/math], which has the same value.)

[code]%3E%3E%3E pprint.pprint( [ "".join( x ) for x in itertools.product( "NE", repeat=8 ) if x.count( "N" ) == 5 and x.count( "E" ) == 3 ] )
['NNNNNEEE',
 'NNNNENEE',
 'NNNNEENE',
 'NNNNEEEN',
 'NNNENNEE',
 'NNNENENE',
 'NNNENEEN',
 'NNNEENNE',
 'NNNEENEN',
 'NNNEEENN',
 'NNENNNEE',
 'NNENNENE',
 'NNENNEEN',
 'NNENENNE',
 'NNENENEN',
 'NNENEENN',
 'NNEENNNE',
 'NNEENNEN',
 'NNEENENN',
 'NNEEENNN',
 'NENNNNEE',
 'NENNNENE',
 'NENNNEEN',
 'NENNENNE',
 'NENNENEN',
 'NENNEENN',
 'NENENNNE',
 'NENENNEN',
 'NENENENN',
 'NENEENNN',
 'NEENNNNE',
 'NEENNNEN',
 'NEENNENN',
 'NEENENNN',
 'NEEENNNN',
 'ENNNNNEE',
 'ENNNNENE',
 'ENNNNEEN',
 'ENNNENNE',
 'ENNNENEN',
 'ENNNEENN',
 'ENNENNNE',
 'ENNENNEN',
 'ENNENENN',
 'ENNEENNN',
 'ENENNNNE',
 'ENENNNEN',
 'ENENNENN',
 'ENENENNN',
 'ENEENNNN',
 'EENNNNNE',
 'EENNNNEN',
 'EENNNENN',
 'EENNENNN',
 'EENENNNN',
 'EEENNNNN']
[/code]Or graphically:

How many balls are in a triangular pyramid with 6 balls on each edge? It’s [math]{6 + 2 \choose 3} = 56[/math]. (See the OEIS demo using this sequence [ https://oeis.org/demo3.html ].) You get another solution here, which is “what is the sum of the first 6 triangular numbers?”

What is [math]\displaystyle \sum_{1 \leq i \leq j \leq 7} | i - j |[/math] ?

[code]%3E%3E%3E sum( ( abs( i - j ) for j in range( 1, 8 ) for i in range (1, j+1) ) )
56
[/code](That came from the OEIS entry on Tetrahedral numbers, and there are many more applications there: A000292 - OEIS [ https://oeis.org/A000292 ])

Once you know to compute the binomial coefficients, they can solve an extremely broad range of problems.

Dhyey Rajyaguru · Answer

I think I am eligible to answer this question as when I was 17, I used to compare myself with others.

Firstly, you need to understand that it is a basic human nature to compare,moreover the idea and mindset that the society has(specially in academics) conditions a kid to constantly compare with others.Comparison is not bad. It actually helps you to improve.

The problem is comparing yourself with others.This is because when you compare yourself with others,you only see the results not the efforts.

So,when you look at a person who is better than you,there is a feeling of dejection and when you look at a person who is not good compared to you,there you feel elated.Both of the cases hinders your growth.

Image courtesy:Google

The above image explaines the problem.

The solution is to compare with the person you know the most i.e. YOU. It will help you to improve everyday,to be the better version of yourself.

So now onwards, set a goal for yourself and strive to achieve it. Don't think about what others have achieved, just focus on your process.Comparison can be the best tool for you to achieve greatness if you use it in the best way.

All the best :)

If you have any problem,feel free to DM me.

Anonymous · Answer

The question refers to a lock "above", so I'll first add a picture.

You turn the first wheel to get the first digit of the "combination" (it's a bad word in this context, but that's how it's called in terms of combination locks), second one to get the second digit, third one to get the third digit. Just like a PIN at an ATM or your phone, it will unlock only if you got the sequence of numbers right, not the set of numbers (i.e. if your credit card PIN is 1234, 4321 won't work [and contrary to hoaxes and urban legends, it will not alert the police]).

So it's just rather vague wording in the text you are reading. I'd replace

In the lock above, there are 10 numbers to choose from (0,1,...9) and we choose 3 of them.

with

In the lock above, there are three wheels with 10 digits to choose from (0,1,...9) and we choose one digit on the each wheel.

Sankush Gupta · Answer

Ques 1 :

Case 1:  When we take either 'O' or 'A':

First lets select the vowels . We can either select O or A so there are 2 ways. Now selecting the other three letters, there are 4C3 ways to do so. Finally, we permute these selections. There are 4! ways to permute four letters where no letter is being repeated.

So, total no. of ways=2*4C3*4!=2*4*24=192 ways

Case 2: When we select both 'O' and 'A':

First lets select the vowels . We have to select both O and A so there is just one way(2C2=1). Now selecting the other two letters, there are 4C2 ways to do so. Finally, we permute these selections. There are 4! ways to permute four letters where no letter is being repeated.

So, total no. of ways=1*4C2*4!=6*24=144 ways

Therefor the answer to the question= Case1 + Case2=192+144=336

Ques 2:

Let these be the ten seats:

- - - - - - - - - -

First lets select the seat for Alice. As both seats adjacent to Alice should be occupied by her friends, she cannot be seated on the extreme ends of the row. So no. of ways to seat Alice is 8. Now lets select 2 friends of Alice who have to be seated adjacent to her. No. of ways to select 2 friends out of her five friends is 5C2. Now lets seat these two friends. There are 2! ways to do this. Now lets choose seats for the remaining 3 friends of Alice. No. of ways to select 3 seats out of the 7 vacant seats=7C3 . We again have to arrange the 3 friends on the 3 chosen seats in 3! ways.

So total no. of ways=8*5C2*2*7C3*6=33600 ways

I hope this answers your questions and please check the answers if you have them and reply in the comments section. Thanks.

Nick Shales · Answer

There are a few interpretations possible here. I will assume that you require the total number of possible gift allocations given that no couple gets a gift from their spouse, no individual gives themselves a gift and everyone receives exactly 1 gift.

This is then reducible to the problem of counting placements of 8 non-attacking rooks on the following “chess board”

Source: Bluffton University Rook Polynomial Calculator - Darryl Nester (second link below)

Each person A, B, C, D, E, F, G, H gives respective gifts 1, 2, 3, 4, 5, 6, 7, 8 and the couples are A&B, C&D, E&F.

8 Rooks are placed so as to avoid black squares and to allocate gifts 1 to 8 to people A to H.

The standard approach involves determining the rook polynomial [ https://en.m.wikipedia.org/wiki/Rook_polynomial ] (wiki link) for the forbidden subboard (black squares). In this case the modified** rook polynomial [ http://www.bluffton.edu/homepages/facstaff/nesterd/java/rookpolynomials.html ] (polynomial calculator link) is

[math]R(x)=(2-4x+x^2)^3(-1+x)^2=x^{7} + 79 \, x^{6} - 232 \, x^{5} + 386 \, x^{4} - 376 \, x^{3} + 212 \, x^{2} - 64 \, x + 8[/math]

Where the term [math]2-4x+x^2[/math] is the rook polynomial for each of the [math]2	imes 2[/math] subboards and the term [math]1-x[/math] is the rook polynomial for each of the two single black squares at the bottom right. We are allowed to multiply because these subboards are all disjunct.

The desired count is given by replacing the each [math]x^k[/math] with [math]k![/math] in the expansion of [math]R(x)[/math]. This is achieved by means of the fact that [math]\int_0^{\infty} x^ke^{-x}dx=k![/math], hence

[math]	ext{desired count}=\displaystyle\int_0^{\infty} (2-4x+x^2)^3(-1+x)^2e^{-x} dx = 6176\qquad\blacksquare[/math]

**Modified in the sense that each coefficient of [math](-1)^kx^{n-k}[/math] is the number of possible placements of [math]k[/math] non-attacking rooks on the forbidden subboard with smallest length n. The polynomial calculator outputs the standard rook polynomial for the forbidden subboard for which the coefficient of [math]x^k[/math] is the number of possible placements of [math]k[/math] rooks.

Nikhil Gupta · Answer

1)
 * Select one letter from A or O n 2C1 ways.
 * Now select 3 letters from the remaining 4 letters in 4C3 ways.
 * And we have 4! to permutate them.
 * Till now we have not excluded cases involving both A and O.
 * So we choose 2 letters from A and O on 2C2 that is 1 way.
 * And we choose test two letters from remaining 4 letters in 4C2 ways. And we multiply it with 4!

Final answer
2C1 *4C3*4! + 2C2*4C2*4!
= 576

2)
 * Now select 2 out of those 5 who will sit next to her.
 * Now we have a group of three sittings ( Alice and her two friends next to next) and the rest 3 friends.
 * Now we have to make them sit and permutate them as well.
 * You should be able to see that if we select the consecutive seats, then do so we have only 8 ways. So we select any one of these combo of three seats. Also, we will multiply with 2! So that the two friends will interchange there positions with respect to Alice.
 * We select any three seats from the rest 7 in 3C7 ways and permutate them in 3! Ways.

Final answer
33600

Rafal Szymanski · Answer

Suppose you wanted to compute how many ways there are to make $20, and that you number each way to make $10, {A,B,C}. Then, to make $20, you can do:

{A,A}
{A,B}
{A,C}
{B,B}
{B,C}
{C,C}

6 ways total. Notice how you can make $20 in 1 + 2 + 3 different ways (3 ways where A is first, 2 ways there B is first, and 1 way where C is first).
1 + 2 + 3 = 3*4/2 = 6.

So for 201 different $10s, you need the above multisets to have exactly 201 elements. So you'll have 1+2...+202 different ways to do that, which is equal to (202*203)/2 = 20503.

This topic is referred to as multisets - have a look here for some more information: Page on wolfram.com [ http://mathworld.wolfram.com/Multichoose.html ]

Shashank Sharma · Answer

Comparison is a thing which one usually does unaware of the consequences that he/she might have to face. Life is full of ups and down, at a moment one is king another moment he could become a beggar,when no money is left. In comparison the underlying thought or belief is that one is better than other or one is worse than other.

If one feels that one is better than other he comes in ego, he tries to belittle others. Don't value others, looks down upon them. Laugh on them or ridicule them : for what they are. This destroys the relationship to the core,he is not able to sustain good relationship because what he feels is superior and this leads to superiority complex .

On the other hand if one feels that one is lower than others he will always try to belittle himself in front of others seeing others values or achievement he will criticise himself for what he is. Punish himself and cause self inflicting pain. That pain which is so harsh that it dwindles and shakes one self confidence.

So at last there should not be an iota of doubt on saying that “Comparison is one which leads to superiority or inferior complex”

Edward Doolittle · Answer

You can think of the operation of erasing the two numbers [math]x[/math] and [math]y[/math] and replacing them with [math]x + y + xy[/math] as a binary operation something like [math]+[/math] or [math]	imes[/math]. Let’s investigate the properties of that binary operation, which we will call [math]*[/math].

We have [math]x * y = x + y + xy = y + x + yx = y * x[/math], so the operation is commutative.

We have, on the one hand, [math](x*y) * z = (x + y + xy) * z = x + y + xy + z + xz + yz + xyz = x + y + z + xy + yz + zx + xyz[/math]. On the other hand, [math]x*(y*z) = x * (y + z + yz) = x + y + z + yz + xy + xz + xyz[/math]. Altogether, [math](x*y)*z = x*(y*z)[/math], so the operation is associative.

The commutative and associative properties together guarantee that when we apply the operation on a pile of numbers, the final result doesn’t depend on the order in which we apply the operation. Think about all the different ways of applying the operation to four numbers, for example: [math]x*(y*(z*w)) = (x * y)* (z * w) = (x*(y*z))*w[/math] etc., etc., just by the associative property; when we involve the commutative property as well, the order of the letters will change, but the final results will all be equal.

Picking two numbers and applying the operation is like making those two numbers the innermost bracket in a large expression like the one in the previous paragraph. In the end, there will be 100 terms, but no matter how the expression is formed, the final result will be the same.

So there is only one possible answer.

Figuring out what that answer will be is another hard problem. The issue is that the answer will be huge. I’ll let you think about it a bit, but here’s a hint: [math]x + y + xy = (1+x)(1+y)-1[/math].

Diego Jancic · Answer

I'm not going to resolve it here but I'll tell you what approach I would take.

Let's say m=5 so we need to find numbers with 5 ones. The smallest one is always going to be the one with no zeros 11111.

The next number is going to be the one with one zero after the first 1, 101111. The third number on the list will be 110111. You can see that the zero is moving to the right.

After all those numbers with one zero you will have he ones with 2 zeros. 1001111. You need to move the zeros to make all possible combinations: 1010111, 1011011, 1011101, 1011110. Then moving the other zero: 1100111, 1101011, 1101101, 1101110. You get the idea.

Now, you could try all combinations until you get the number you want or fast forward to the number you want.

If m=5, and the are zero 0s, you have only one combination possible, with one zero you have 5 combinations, with 2 zeros you have 15 combinations... Find the sequence and problem solved!

Anonymous · Answer

Try this method.

1) The number of ways you can select any two sections is given by 3c2
= 3

2) Now, focus on the sections. From each section at least two questions must be done. Since you are choosing from 5 questions in each section, the number of ways you can do this is:
(2,5), (3,4), (4,3), (5,2)

= 5c2*5c5 + 5c3*5c4 + 5c4*5c3 + 5c5*5c2
= 2(10 + 50)
= 12o

3) Now, multiply the result in (2) with (1), since there are 3c2 ways you can do (2)
3c2 * (5c2*5c5 + 5c3*5c4 + 5c4*5c3 + 5c5*5c2)
= 3*120
= 360

Well, I think the fault lies in the fact in this step:

%3E Number of ways of selecting at least 2 questions from each section 5c2....

%3E Then, we have to select 3 questions from the remaining 6: 6c3 = 20

Though I am not sure of the reasoning, I am certain the fault lies in this step.

Anonymous · Answer

I think you question is probably incomplete or copied wrong.
Anyways, considering this problem, it is actually very easy to select 8 books from the set of 36. 
consider the set of all books i.e , the set contains 36 books.
You are given a job to show , in how many ways can you select 8 books, probability of each book being uniformly distributed.
Now you are given n sets of 8. You need to figure out n.
Consider each place in the set , if you start with the first place from the 8 places you can have 36 in the first place. Now that you have selected one book from the 36 books, you are left with 35 options of books to be selected. Clearly you can see the pattern i.e for 8 places , after each place there would be one less possible books to be selected from the previous one.

Mathematically writing this :- 
36*35*34*33*32*31*30*29*28 possibilities.
Now rather that using this pattern you can generalize any problem by using the factorial function. 
36! = 36*35*34*....*1!*0!.
but you need only first 8 elements of this 36 set.
therefore 36!/(36-8)! is the ans.

Taha Nakabi · Answer

The number of combinations without taking in account the constraint is [math]C[/math](8,2)=28 now we should eliminate the number of combinations of the three couples which will give you 28–3=25

P.S [math]C[/math](n,k)[math]=n!/(n-k)!k![/math]

Laura Bean · Answer

My original thought is: check and guess. It must be 1,2 or 2,1. So check and guess x=1, y = 2. 
Substituting x for 1 and y fir 2, 
You have: 
1+2 +(1×2)=(1+1)(1+2) -1
3+2=2(3)-1
5=6-1
5=5
So an original assumption is right.,x=1 , y=2.

Rory Fellowes · Answer

If it is inn answer to a statement or question, it would be all right. For example: “Can you run fast?” “Faster than you!” would work (though itwould be a bit rude!). But if it is said without any precursive question, then the person addressed would probably ask, “What is faster than me?” The same would apply to “More than ten” - “How many are there?” “More than ten.” Otherwise, you would be asked “What is more than ten?”

In other words, the first member is necessary, but it may be supplied by a precursive statement or question.