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Re: In how many ways can 12 different books be distributed equally among 4
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24 Nov 2018, 10:15
1
Books have to be divided equally. So each box should get 12/4 = 3 books only. So 1st box gets 3 books from 12 in 12C3 ways. Now, 9 books remaining. So 2nd box gets 3 books from 9 in 9C3 ways. Now 6 books remaining. So 3rd box gets 3 books from 6 in 6C3 ways. Now 3 books remaining. So 4th box gets 3 books from 3 left in 3C3 ways or 1 way.
Ways of distributing 12 books among 4 similar boxes will be 12C3*9C3*6C3*1 Assuming that different boxes means which books go into which box will make a difference, then we will be able to arrange the total ways of distributing 12 books in 4! ways.
In how many ways can 12 different books be distributed equally among 4 different boxes?
A) 12C3 B) 12C4 C) 12C3*9C3*6C3 D) 12C4*8C4 E) 12C3*9C3*6C3*4!
12 different books be distributed equally among 4 different boxed in 12C3*9C3*6C3 ways Ways to arrange these books in 4 different boxes is 4! So the answer is 12C3*9C3*6C3*4!
Re: In how many ways can 12 different books be distributed equally among 4
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24 Nov 2018, 14:54
For distributing 12 books equally across 4 boxes, we need to split 3 books in each.
First 3 books can be selected in 12C3 ways, second 3 is 9C3 ways, third 3 in 6C3 ways and last 3 in 3C3 ways.
Now, let four boxes be A, B, C and D. First box can be selected in 4 ways as it can be any of A, B, C or D. Lets say A is selected as first box. Similarly, Second box can be selected in 3 ways as it can be any from B, C and D and third box can be selected in 2 ways and last box in 1 way.
So, total number of ways = 12C3*9C3*6C3*3C3*4*3*2*1 = 12C3*9C3*6C3*1*4!
In how many ways can 12 different books be distributed equally among 4 different boxes?
A) 12C3 B) 12C4 C) 12C3*9C3*6C3 D) 12C4*8C4 E) 12C3*9C3*6C3*4!
Take the task of distributing the books and break it into stages.
We must place 3 books in each of the 4 boxes. So, let's call the boxes box #1, box #2, box #3 and box #4
Stage 1: Select 3 books to go in box #1 Since the order in which we select the books does not matter, we can use combinations. We can select 3 books from 12 books in 12C3 ways So, we can complete stage 1 in 12C3 ways
Stage 2: Select 3 books to go in box #2 There are 9 boxes remaining. So, we can complete this stage in 9C3 ways
Stage 3: Select 3 books to go in box #3 There are 6 boxes remaining. So, we can complete this stage in 6C3 ways
Stage 4: Select 3 books to go in box #4 There are 3 boxes remaining. So, we can complete this stage in 3C3 ways
By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 12 books) in (12C3)(9C3)(6C3)(3C3) ways
Check the answer choices....our answer doesn't seem to be there. However, if we recognize that 3C3 = 1, we can see that [color=blue(12C3)(9C3)(6C3)(3C3)[/color] = (12C3)(9C3)(6C3)(1) = (12C3)(9C3)(6C3)
Answer: C
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.