3. Gummi bears come in twelve flavors, and you have one of each flavor Answer ea
ID: 3251585 • Letter: 3
Question
3. 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) (2 points) 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).Explanation / Answer
Solution:
(a)
First let us select 3 from 12 for the first row in 12C3 = 220 ways
then we select 4 from remaining 9 in 9C4 = 126 ways
and finally we have 5 left and we can select 5 in 5C5 = 1 ways
Thus, we have total of (220 × 126 × 1) = 27720 ways
(b)
This is a 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.
Therefore, the numbers of ways of arranginging n people in a circle is (n-1)!.
Therefore 12 bears can be arranged in a circle in (12-1)! = 11! ways = 39916800 ways
Again, according to the given question, 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 39916800 ways/2 = 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 . There are total 11 gaps.
We have to put 2 seperators in any two gaps (this will form 3 groups) out of 11 gaps
We can do this by 11C2 = 55 ways
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