johnwesley wrote:
How many ways can the letters in the word COMMON be arranged?
A. 6
B. 30
C. 90
D. 120
E. 180
This is a permutation with indistinguishable events - repeated items. The number of different permutations of N objects, where there are N1 indistinguishable objects of style 1, N2 indistinguishable objects of style 2, ..., and Nk indistinguishable objects of style k, is = N!/(N1!*N2!* ... * Nk!). In this case, N=6; N1=2, and N2=2. This gives the formula: 6!/(2!*2!)=180
THEORY FOR SUCH KIND OF PERMUTATION QUESTIONS: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 \(\frac{9!}{4!3!2!}\).
BACK TO THE ORIGINAL QUESTION:How many ways can the letters in the word COMMON be arranged?A. 6
B. 30
C. 90
D. 120
E. 180
According to the above the # of permutations of 6 letters COMMON out of which 2 O's and 2 M's are identical is \(\frac{6!}{2!*2!}=180\).
Answer: E.
Hope it's clear.
I get the idea behind the formula with repeats but shouldn't we be multiplying 180 x 4 here? O and M can each be flipped in terms of their respective orders?
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