How To Understand Combinations Using Multiplication

Multiplication is a wonderful little operation. Depending on the context, it can

And today we’ll see yet another use: listing combinations.

This is a companion discussion topic for the original entry at

Excellent discussion. I would also add the example of genetics. A woman(XX) mates with a man (XY) to achieve (XX + XY + YX + XX) = (2XX + 2XY)–what do you know, equal odds of a boy or girl.

I remember the trick was useful during my high school biology class when we had to cross 4 traits (say AaBc crossed with aabC). Others started drawing out Punnet squares, but I used the algebraic notation plus a symbolic calculator.

Nice explanation. Your figure are really nice. Which program you use to draw those figures?

@Presh: That’s a great use of this trick! I had completely forgotten about mixing genes in biology, but that’s exactly it. The child gets (x OR x) AND (x or y) = (x + x)*(x+y) = 2xx + 2xy, or an equal chance as you say.

@sgwong: Glad you liked it – I use PowerPoint 2007 to make the diagrams.

I came to your website after a while Kalid. And the only word that came to mind was “COOL”. Was very busy during the past month on my job so it was a great refresher coming back and brushing up my combinations and probability :)…am really looking forward to more of your math posts that have once again sparked my interest in math after my university professors had managed to kill it. In the meanwhile I will be treating myself to some of your other posts which I have missed over the course of the past month.

Your fan from Pakistan

Hi Mohammad, thanks for the wonderful note! Glad you’re enjoying the articles.

Hi Khalid

You are real thinker! Trying to study math again,after 10 yrs, at college is a real nightmare! I fell in love with math again after 10 yrs after reading you posts! You have an amazing talent to dissect and disseminate what looks like complex theories.

Thank you very much for sharing your thoughts. Its not just about the content of the posts, but a whole new way of thinking that you opened up.

Best Regards

Thanks for the kind words Meera! Really glad you were able to rekindle your interest :).

[…] Following on from recent posts about expanding brackets and the binomial expansion, betterexplained uses the same principles to discuss combinations (a key component of the binomial expansion when you’re using positive integer powers). I think this is an interesting way of seeing how you work out nCr. […]

I see that the co-efficient in the expansion gives the number of options of re-arranging. Doesn’t this seem like a permutation rather than a combination? Isn’t it giving us all possibilities of re-arranging? Yet it yields the same value as combination. (ex: how many ways to pick one from 3 = coefficient in 2nd term in the expansion of (1+x)^3) I just can’t understand why they are the same number when one seems like permutation and the other a combination?? Any help would be amazing. That was a great read by the way. Thank you

@Jaq: Great question – the expansions for the Binomial Theorem can be tricky.

Here’s one way to think about it. For (a+b)^10, it means:

Choose a or b, 10 times.

Suppose we want to know the coefficient for a^3 * b^7. The neat thing is that picking 3 elements from 10 to “become a’s” means the other 7 automatically become b’s. They have to, since we want a^3*b^7; every element is used.

So how many ways can we pick 3 items from 10? At first we’d think 1098 – a permutation.

But because we’re multiplying, the order doesn’t matter: Saying element 1, 2, and 3 become a’s (in that order) is the same as saying element 3, 2 and 1 become a’s in that order.

So, we do C(10,3) which is the number of groups of 3 we can pick from 10. Permutations would be important if the order of multiplication mattered, but they don’t in this case so we have to divide by the redundancies.

Hope that helps!

I have a question Sir -

I have 10 numbers for example - 1,2,3, 4,5, 6,7,8,9,10 arranged in two sets of five numbers each

I arrange the numbers – in the following format

(1 2 3 4 5 ) and ( 6 7 8 9 10)

A = 1 , A = 6
B= 2 , B= 7
C=3 , C = 8
D= 4 , D= 9
E = 5 , E = 10

TASK – Arrange the numbers into groups of five making sure that EACH combination of five numbers reflects ABCDE

NOTE - The format ABCDE must be strictly adhered to when arranging the numbers in groups of five

Some examples are – (1,2,3,4,5) , (6,7,8,9, 10), ( 1,2,3,9,10) (6, 7, 8, 4,5)


COULD THIS BE DONE FOR 16 numbers arranged in 2 sets of eight and how would that case be solved ?

i recieved very nice explanation on permutation and combination

I can’t seem to understand how to solve a problem similar to the following:

If there are X different items and I need to determine the number of subgroups containing at least Y of these items, how would I go about doing that? The actual problem asks for the above with x=6 and y=2, and the answer is 57 b/c they do 2^6 - 6 - 1. But why do they do that? I don’t understand, please help!!

Many thanks,

I am still confused with permutation. My son has a math problem -
How many different arrangements can be made with the letters in the word MOVIE?
I don’t see how they get the answer.

Really helpful…m very bad at permutations and combinations like difficult problems in GMAT wher you have 5 married couples and you pick 3 people out .what is the prob.niether is married to each other.can you help me an easy way to silve these problems.will be really much of help.Thanks in advance

[…] Navigate a Grid Using Combinations And Permutations […]

What would the number of combinations be to the two following questions…
*… There are 36 two sided objects. Every side is unique in pattern. If they remain in the same static sequence how many different combination of sides are there?
*… There are 49 cubes. Every side of all the cubes is unique in pattern. So there are 294 unique patterns in the set. Each cube can be re-arranged in the sequence of the 49 cubes and each side of every cube can be face-up. How many different combinations of the 49 cubes can be created using all sides of the cubes?

This is really great for me it is helpful to me. Thanks to the group that came up with this it is great really!!!

@Brianna: Awesome, glad it helped!