Hi! This is a question that my daughter receivied; however, I cannot remember the calculation as Math was so long ago for me... :confused:

If there are 54 numbers and you have to group them in 6's, how many different sequences can all 54 numbers be placed? *** You can use numbers repeated, but not in the same sequence.

From what I think I remember, 6*54=324, but this just doesn't seem right to me...



Any help provided is appreciated as I am lost... Thanks!;)

I'm assuming that order of elements in a set does not matter for this problem. (1,2,3 is the same as 3,2,1)

From: Combination - Wikipedia, the free encyclopedia

Quote:

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The number of k-combinations (each of size k) from a set S with n elements (size n) is the binomial coefficient

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If you forget, n! Is the product of all positive integers less than or equal to n. So 5!=5*4*3*2*1=120

So, for your daughter's problem, the set has 54 elements so n becomes 54 and you want to select groups of 6 so k becomes 6.


Now I'm sure you're saying, "great, how do I calculate 54!?", or at least you should be, because 54! Is a pretty large number. (it is about 72 digits long) but imagine the following.
54!=54*53*52*51*50*49*48*47*... *2*1
and
48!=48*47*... *2*1
so all the 48*47*... *2*1 from 54! Will cancel with all the terms from 48! So you've just got 54*53*52*51*50*49 on the top and 6! On the bottom.
I don't think there's a really simple way to calculate 54*53*52*51*50*49 but it's not too bad that you couldn't do it by hand. Many calculators can't do this math because it takes more digits than they have. You could use a calculator to multiply most of the terms and then do the last few by hand, or use the calculator in windows, it can do bigger math. The answer is 18,595,558,800.
And 6! Is easy, just 720. So now you have 18,595,558,800/720. You can do this math by hand also if needed. Or just use windows calculator again. 18,595,558,800/720 = 25,827,165 so there are 25,827,165 ways of picking 6 numbers out of a list of 54 numbers.

I hope this helps. Please let us know if you have any questions.