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few permutation questions

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by fangtray » Mon Sep 19, 2011 8:49 pm
1. in how many ways can a committee of 2 boys and 3 girls can be made out of 5 boys and 6 girls?


2.there are 4 copies of 5 different books. in how many ways can they be arranged on a shelf?

3. in how many ways a three digit number can be written using all of the digits from 1 to 9 without repetition of any digit?

4. in how many ways can the letters of the word "missisippi" be arranged?


sorry guys i'm very bad at permutations and combinations... instead of just providing something like 7*6*5*4*3 can someone explain why they do it that way?

thanks for all of your help
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by Anurag@Gurome » Mon Sep 19, 2011 9:10 pm
In how many ways can a committee of 2 boys and 3 girls can be made out of 5 boys and 6 girls?
2 boys can be selected from 5 boys in 5C2 ways.
3 girls can be selected from 6 girls in 6C3 ways.
Therefore, required number of committees = 5C2 * 6C3 = 5!/(3! * 2!) * 6!/(3! * 3!) = 10 * 20 = 200
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by Anurag@Gurome » Mon Sep 19, 2011 9:17 pm
There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?
4 copies of 5 different books means in all there are 4 * 5 = 20 books, which can be arranged on the shelf in 20! ways
Therefore, no. of ways in which 4 copies of 5 different books can be arranged on a shelf = [spoiler]20!/(4!)^5[/spoiler]
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by Anurag@Gurome » Mon Sep 19, 2011 9:25 pm
In how many ways can the letters of the word "missisippi" be arranged?
No. of letters in the word "missisippi" = 10, which can be arranged in 10! ways
Now, no. of "s" = 3
No. of "i" = 4
No. of "p" = 2
Since the above letters are repeating, so we need to divide these repetitions.

Required no. of ways = 10!/(3! * 4! * 2!) = 10 * 9 * 8 * 7 * 6 * 5/(3! * 2!) = 10 * 9 * 4 * 7 * 5 = 12,600
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by LalaB » Mon Sep 19, 2011 9:45 pm
fangtray wrote:
3. in how many ways a three digit number can be written using all of the digits from 1 to 9 without repetition of any digit?
9*8*7
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by LalaB » Mon Sep 19, 2011 9:48 pm
Anurag@Gurome wrote:
There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?
4 copies of 5 different books means in all there are 4 * 5 = 20 books, which can be arranged on the shelf in 20! ways
Therefore, no. of ways in which 4 copies of 5 different books can be arranged on a shelf = [spoiler]20!/(4!)^5[/spoiler]
I thought the question asks all books arrangement, and got the answer 20!
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by fangtray » Tue Sep 20, 2011 2:44 am
LalaB wrote:
fangtray wrote:
3. in how many ways a three digit number can be written using all of the digits from 1 to 9 without repetition of any digit?
9*8*7
thats what i thought at first. but its actually 9^3 ways. at least thats what is on the answer key. does somebody know?
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by krishnasty » Tue Sep 20, 2011 3:38 am
fangtray wrote:
LalaB wrote:
fangtray wrote:
3. in how many ways a three digit number can be written using all of the digits from 1 to 9 without repetition of any digit?
9*8*7
thats what i thought at first. but its actually 9^3 ways. at least thats what is on the answer key. does somebody know?
It cant be 9^3..
The first digit can be filled by any number from 1-9...9 ways
the scond digit can be filled by any number EXCEPT the number chosen for the first digit..hence, 8 ways
similarly, for the 3rd digit, it has to be filled in 7 ways ( coz 2 numbers are already chosen).
total ways : 9*8*7

had it been that the number can be chosen with repetation too, then it would have been 9*9*9..
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by fangtray » Tue Sep 20, 2011 3:47 am
Anurag@Gurome wrote:
There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?
4 copies of 5 different books means in all there are 4 * 5 = 20 books, which can be arranged on the shelf in 20! ways
Therefore, no. of ways in which 4 copies of 5 different books can be arranged on a shelf = [spoiler]20!/(4!)^5[/spoiler]
why divide by 4!)^5?
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by sl750 » Tue Sep 20, 2011 6:13 am
fangtray wrote:
Anurag@Gurome wrote:
There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?
4 copies of 5 different books means in all there are 4 * 5 = 20 books, which can be arranged on the shelf in 20! ways
Therefore, no. of ways in which 4 copies of 5 different books can be arranged on a shelf = [spoiler]20!/(4!)^5[/spoiler]
why divide by 4!)^5?
I believe that is necessary to avoid repetition. It is like the missippii problem posted by you.
The four copies are identical. Since there are 5 different books each book has 4 copies, we divide
by 4!^5
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