Gummi bears come in twelve flavors, and you have one of each flavor. Answer each

Question

Gummi bears come in twelve flavors, and you have one of each flavor. Answer each question and briefly explain your reasoning. Leave your answer unsimplified in terms of permutations, combinations, factorials, exponents, etc. (a) How many ways are there to arrange your gummi bears for a picture so that three gummi bears are in the front row, four are in the middle row, and five are in the back row? Two arrangements are considered the same if each gummi bear is in the same row in both arrangements (the order within the rows does not matter). (b) How many ways are there to arrange your gummi bears in a circle? Two arrangements are considered the same if each gummi bear has the same left neighbor and the same right neighbor in both arrangements. (c) How many ways are there to split your Twelve gummi bears among three people so that each of the three people gets at least one gummi bear?

Explanation / Answer

3
a.

This is a problem of selection (and not arrangement) so we shall be using Combination (and not permutation) to solve this:

First let us select 3 from 12 for the first row in 12C3 (or 220) ways
then we select 4 from remaining 9 in 9C4 (or 126) ways
and finally we have 5 left and we can select 5 in 5C5 (or 1) ways

so we have total of (220*126*1) 27720 ways

b.
This is a typical problem of permutation. For circular permutation we need to fix 1 bear on the circumferance of the circle, that does not changes its position. Now the circlular structure can be unrolled to for a linear structure with the fixed bear at one particular end. So the problem now boils down to linear permutation with 11 bears. So the total number of possible arrangements are 11! (or 39916800) ways.
Now the question states that two arrangements are similar if neighbours are swapped i.e abc = cba. So we have to rule out half of the permutaions we predicted since for all permutaions we will have a reverse permutaion.
so the total ways of arrangements are 19958400 ways

c.
Now we are left with 12 bears.
let them be
    1_2_3_4_5_6_7_8_9_10_11_12
let us also have 2 seperators
The problem now is to put 2 seperators in any two gaps (this will form 3 groups) out of 11 gaps
We can do this by 11C2 (or 55)ways

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