[Math] Calculating the number of arrangements of items into compartments

combinatoricspermutations

I'm dealing with intro combinatorics here, and am a bit perplexed by the logic around arranging items in boxes. When arranging a small set of objects in a series of boxes, I am not seeing why the arrangement in the boxes reduces the number of total arrangements.

In looking at this, I am given the formula of:

n!/(n1! x ... x nk!), where n1 + ... + nk = n

For example, the ways of arranging 3 balls in a row is: 3! = 6. This is interpreted as the total number of arrangements n!/(n-k)! where k=3 is the number of objects permuted for n=3 objects in total, so for balls A, B, and C, we get:

ABC
ACB
BAC
BCA
CAB
CBA

However, if we have two boxes available, Box A that will hold two balls, and Box B that will hold only one ball, the ways of arranging them are:

Box A   Box B
AB      C
AC      B
BA      C
BC      A
CA      B
CB      A

This appears to be 6 arrangements, assuming AB and BA are considered different arrangements; however, when plugged into the formula I get:

3!/(2!x1!) = 3*2*1/(2*1x1) = 3

Why does this equation come up with half of the expected arrangements?

Best Answer

The best thing you can do in the second case is, find the number of ways you can choose 2 balls out of 3(to put in the bigger box) and multiply it with the number of ways you can choose 1 ball out of 1 i.e 1 The answer'll be 3C2*1C1=3. The question doesn't require you to arrange the two balls in the bigger box in a specific order whereas in the first case it's like arranging 3 different digits in 3 places.In that case you must take order into account.

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