In how many ways can 5 red,4 blue,and 1 green | vedclass

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Question

(A) The total number of balls is $5 + 4 + 1 = 10$.Since there are $5$ red balls,$4$ blue balls,and $1$ green ball,the number of distinct arrangements

Vedclass · Accepted Answer

$1260$

Vedclass · Answer

(A) The total number of balls is $5 + 4 + 1 = 10$.Since there are $5$ red balls,$4$ blue balls,and $1$ green ball,the number of distinct arrangements is given by the formula for permutations of a multiset:$\frac{n!}{n_1! 	imes n_2! 	imes n_3!} = \frac{10!}{5! 	imes 4! 	imes 1!}$$= \frac{10 	imes 9 	imes 8 	imes 7 	imes 6 	imes 5!}{5! 	imes (4 	imes 3 	imes 2 	imes 1) 	imes 1}$$= \frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{24}$$= 10 	imes 9 	imes 7 	imes 2 = 1260$.

In how many ways can $5$ red,$4$ blue,and $1$ green ball be arranged in a row?

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