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PR1012 - Permutations/combinations 7

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PR1012 - Permutations/combinations 7 [#permalink] New post 20 Dec 2009, 13:15
00:00
A
B
C
D
E

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(N/A)

Question Stats:

50% (02:06) correct 50% (02:44) wrong based on 5 sessions
Another question from PR1012. In this case I think that the stem is way too vague in stating the question. I think it is not clear whether the order of boys and girls matters. Please let me know what you think.

Six kids will be lined up from left to right. Boys may not be at either end (the first and last spot). In how many ways can the kids be lined up?

I) there are 4 girls

II) There are 12 different ways to fill the first and last spot
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Re: PR1012 - Permutations/combinations 7 [#permalink] New post 20 Dec 2009, 21:06
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Well yes it matters, we are not talking about identical white and black balls :)

Now the solution

We know that the total number of kids = 6
Since boys can stand in the corners, we need to know specifically how many boys and girls are there.
Statement 1
If number of girls is 4 then number of boys = 2
This is sufficient to calculate the number of arrangements.
(Not required but I will go ahead and get the number of arrangements)
First we fill in the corners with 2 girls, can be done in 4*3 ways
Now we can arrange the remaining 4 kids in any order = !4
Total number of arrangements = 4*3*!4

Statement 2
there are 12 different ways to fill first and last spot
Let number of girls be n then number of ways to fill the corners= n*(n-1)
n*(n-1)=12
n=-3 or n=4
4 is the only possible value, we can solve as above to get the arrangements.

Answer is D.
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Re: PR1012 - Permutations/combinations 7 [#permalink] New post 22 Dec 2009, 07:44
@atish please explain the funda of 4*3 for the girls.
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Re: PR1012 - Permutations/combinations 7 [#permalink] New post 22 Dec 2009, 08:05
sagarsabnis wrote:
@atish please explain the funda of 4*3 for the girls.

Sure -
If there were 4 girls, and if you needed an arrangement as below

G X X X X G
where the G represents girls (since boys can't stand on the corners) and Xs represent the remaining kids.

The first G can be filled in 4 ways AND the other G can be filled in 3 ways.
Hence the corners can be filled in 4*3 ways.
We have 2 boys and 2 girls to fill the remaining 4 spots - can be done in !4 ways.
Hence the answer is 4*3*!4
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Re: PR1012 - Permutations/combinations 7 [#permalink] New post 22 Dec 2009, 10:40
eehhhhh...why i cant think that way...anyways thanks a lot...
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Re: PR1012 - Permutations/combinations 7 [#permalink] New post 22 Dec 2009, 11:14
sagarsabnis wrote:
eehhhhh...why i cant think that way...anyways thanks a lot...

Comes with practice, you will get there I am sure.
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Re: PR1012 - Permutations/combinations 7   [#permalink] 22 Dec 2009, 11:14
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PR1012 - Permutations/combinations 7

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