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In how many ways can 3 red marbles, 2 blue marbles, and 5 ye
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09 May 2019, 01:20
09 May 2019, 01:20
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Question Stats:
79% (00:33) correct
20% (00:52) wrong based on 29 sessions
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In how many ways can 3 red marbles, 2 blue marbles, and 5 yellow marbles be placed in a row?
(A) \(3!*2!*5!\)
(B) \(\frac{12!}{10!}\)
(C) \(\frac{10!}{3!} * \frac{10!}{2!} * \frac{10!}{5!}\)
(D) \(\frac{10!}{3!*2!*5!}\)
(E) \(\frac{10!}{(3!*2!*5!)^2}\)
(A) \(3!*2!*5!\)
(B) \(\frac{12!}{10!}\)
(C) \(\frac{10!}{3!} * \frac{10!}{2!} * \frac{10!}{5!}\)
(D) \(\frac{10!}{3!*2!*5!}\)
(E) \(\frac{10!}{(3!*2!*5!)^2}\)
Re: In how many ways can 3 red marbles, 2 blue marbles, and 5 ye
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10 May 2019, 02:34
10 May 2019, 02:34
2
It might seem that option A is the correct option however 3 red marbles, 2 Blue marbles and 5 yellow marbles are common interchanging 2 red marbles or 4 yellow marbles will not alter the combination. For eg. Let the two blue marbles be denoted as b1 and b2
Hence an arrangement such that RRRB1B2YYYYY and RRRB2B1YYYYY is the same.
Therefore the combination is given by \(\frac{10!}{3!*2!*5!}\)
There are 10 marbles in all and among them 3red, 2blue and 5yellow
Hence an arrangement such that RRRB1B2YYYYY and RRRB2B1YYYYY is the same.
Therefore the combination is given by \(\frac{10!}{3!*2!*5!}\)
There are 10 marbles in all and among them 3red, 2blue and 5yellow
Re: In how many ways can 3 red marbles, 2 blue marbles, and 5 ye
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31 Aug 2022, 06:40
31 Aug 2022, 06:40
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