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27 Oct 2015, 21:53
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happyface101
THEORY:
Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).
For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).
For more check the links below:
Combinatorics Made Easy!
Theory on Combinations
DS questions on Combinations
PS questions on Combinations
Tough and tricky questions on Combinations
02 Nov 2015, 06:34
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A 4 letter code without constraint i.e, with 4 unique letters can be formed in 4! ways.
Constraint is only 3 letters are unique and one of them repeats.
With constraint the answer is 4! /2! =12 ways
There are three such cases i.e., A ,B or C may repeat.
So the total number of ways = 12*3=36
Constraint is only 3 letters are unique and one of them repeats.
With constraint the answer is 4! /2! =12 ways
There are three such cases i.e., A ,B or C may repeat.
So the total number of ways = 12*3=36
16 May 2017, 18:16
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GHIBI
We are given three letters, A, B, and C, and we must create a four-letter code in which all three letters are used. So, one letter must be repeated. Thus, we have the following three options:
1) A-B-C-A (if A is repeated)
2) A-B-C-B (if B is repeated)
3) A-B-C-C (if C is repeated)
Let’s start with option 1:
We see that there are four total letters and two repeated As. Thus, that code can be selected in the following number of ways:
4!/2! = (4 x 3 x 2 x 1)/(2 x 1) = 4 x 3 = 12 ways
Since the second code, A-B-C-B, has two Bs rather than two As, we can create the second code in 12 ways. Likewise, since the third code, A-B-C-C, has two Cs rather than two As or two Bs, we can create the third code in 12 ways.
Thus, the code can be created in 12 + 12 + 12 = 36 ways.
Answer: C
