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16 Sep 2014, 00:40
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In how many ways can 3 identical red chips and 2 identical blue chips be arranged in a row to form distinct color sequences?
A. 10
B. 12
C. 24
D. 60
E. 100
A. 10
B. 12
C. 24
D. 60
E. 100
16 Sep 2014, 00:40
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Bookmarks
Official Solution:
In how many ways can 3 identical red chips and 2 identical blue chips be arranged in a row to form distinct color sequences?
A. 10
B. 12
C. 24
D. 60
E. 100
THEORY
The number of permutations for \(n\) objects, where \(P_1\) objects are identical of one kind, \(P_2\) are identical of a second kind, \(P_3\) are identical of a third kind, ... , and \(P_r\) are identical of the \(r^{th}\) kind such that \(P_1 + P_2 + P_3 + ... + P_r = n\), is given by:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\)
For instance, the number of permutations for the letters in the word "gmatclub" is \(8!\), since all 8 letters are distinct.
The number of permutations for the letters in the word "google" is \(\frac{6!}{2! *2!}\), since out of the 6 letters, "g" and "o" each appear twice.
The number of permutations for 9 balls, with 4 being red, 3 green, and 2 blue, is \(\frac{9!}{4! *3!*2!}\).
BACK TO THE QUESTION
In our case, the permutations for 5 letters, BBRRR, where there are 2 identical Bs and 3 identical Rs, is \(\frac{5!}{2! *3!} = 10\).
Answer: A
In how many ways can 3 identical red chips and 2 identical blue chips be arranged in a row to form distinct color sequences?
A. 10
B. 12
C. 24
D. 60
E. 100
THEORY
The number of permutations for \(n\) objects, where \(P_1\) objects are identical of one kind, \(P_2\) are identical of a second kind, \(P_3\) are identical of a third kind, ... , and \(P_r\) are identical of the \(r^{th}\) kind such that \(P_1 + P_2 + P_3 + ... + P_r = n\), is given by:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\)
For instance, the number of permutations for the letters in the word "gmatclub" is \(8!\), since all 8 letters are distinct.
The number of permutations for the letters in the word "google" is \(\frac{6!}{2! *2!}\), since out of the 6 letters, "g" and "o" each appear twice.
The number of permutations for 9 balls, with 4 being red, 3 green, and 2 blue, is \(\frac{9!}{4! *3!*2!}\).
BACK TO THE QUESTION
In our case, the permutations for 5 letters, BBRRR, where there are 2 identical Bs and 3 identical Rs, is \(\frac{5!}{2! *3!} = 10\).
Answer: A
23 Oct 2023, 01:43
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I've revised the question and solution, incorporating additional details for improved clarity. I trust this makes it more comprehensible.
