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There are 4 boys and 8 girls taking place in a singing competition. The number of different ways in which each boy is paired with a girl partner and the four remaining girls are paired into two pairs each of the two is

Option: 1

2260


Option: 2

2160


Option: 3

3160

 


Option: 4

3260


Answers (1)

Given that,

4 women can be selected in {}^8C_4 ways.

Each girl is paired with 4 boys in 4! Ways.

The remaining 4 girls can be grouped into 2 couples is given by,

\frac{4 !}{2 ! \times 2 ! \times 2 !}=\frac{3}{2}

The number of ways is,

\begin{aligned} & { }^8 C_4 \times 4 ! \times \frac{3}{2}=\frac{8 !}{4 ! 4 !} \times 4 ! \times \frac{3}{2} \\ & { }^8 C_4 \times 4 ! \times \frac{3}{2}=4 \times 7 \times 30 \times 3 \\ & { }^8 C_4 \times 4 ! \times \frac{3}{2}=2160 \end{aligned}

Therefore, the required number of ways is 2160 ways.

Posted by

Kshitij

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