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
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.
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86% (01:30) correct
