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
few permutation questions
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2 boys can be selected from 5 boys in 5C2 ways.In how many ways can a committee of 2 boys and 3 girls can be made out of 5 boys and 6 girls?
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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4 copies of 5 different books means in all there are 4 * 5 = 20 books, which can be arranged on the shelf in 20! waysThere are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?
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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No. of letters in the word "missisippi" = 10, which can be arranged in 10! waysIn how many ways can the letters of the word "missisippi" be arranged?
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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- LalaB
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I thought the question asks all books arrangement, and got the answer 20!Anurag@Gurome wrote:4 copies of 5 different books means in all there are 4 * 5 = 20 books, which can be arranged on the shelf in 20! waysThere are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?
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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thats what i thought at first. but its actually 9^3 ways. at least thats what is on the answer key. does somebody know?LalaB wrote:9*8*7fangtray 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?
- krishnasty
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It cant be 9^3..fangtray wrote:thats what i thought at first. but its actually 9^3 ways. at least thats what is on the answer key. does somebody know?LalaB wrote:9*8*7fangtray 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?
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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why divide by 4!)^5?Anurag@Gurome wrote:4 copies of 5 different books means in all there are 4 * 5 = 20 books, which can be arranged on the shelf in 20! waysThere are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?
Therefore, no. of ways in which 4 copies of 5 different books can be arranged on a shelf = [spoiler]20!/(4!)^5[/spoiler]
- sl750
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I believe that is necessary to avoid repetition. It is like the missippii problem posted by you.fangtray wrote:why divide by 4!)^5?Anurag@Gurome wrote:4 copies of 5 different books means in all there are 4 * 5 = 20 books, which can be arranged on the shelf in 20! waysThere are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?
Therefore, no. of ways in which 4 copies of 5 different books can be arranged on a shelf = [spoiler]20!/(4!)^5[/spoiler]
The four copies are identical. Since there are 5 different books each book has 4 copies, we divide
by 4!^5