There are 3 men and 7 women taking a dance class. Number of different ways in which each man be paired with a woman partner, and the four remaining women be paired into two pairs each of two, is

Question

There are 3 men and 7 women taking a dance class. Number of different ways in which each man be paired with a woman partner, and the four remaining women be paired into two pairs each of two, is (A) 105 (B) 315 (C) 630 (D) 450

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/f0c06351af0c14668da3150898157608-.png" /> 3 women can be selected in 7 C 3 ways and can be paired with 3 men in 3 ! ways. Remaining 4 women can be grouped into two couples in (4 !/2 ! ⋅ 2 ! ⋅ 2 !)=3 ∴ total = 7 C 3 ⋅ 3 ! ⋅ 3=630

Q. There are 3 men and 7 women taking a dance class. Number of different ways in which each man be paired with a woman partner, and the four remaining women be paired into two pairs each of two, is

105

315

630

450

3 women can be selected in $^{7} C_{3}$ ways and can be paired with 3 men in 3 ! ways.
Remaining 4 women can be grouped into two couples in $\frac{4 !}{2 ! \cdot 2 ! \cdot 2 !} = 3$
$∴$ total $=^{7} C_{3} \cdot 3! \cdot 3 = 630$

Q. There are 3 men and 7 women taking a dance class. Number of different ways in which each man be paired with a woman partner, and the four remaining women be paired into two pairs each of two, is

105

315

630

450

3 women can be selected in 7C3​ ways and can be paired with 3 men in 3 ! ways. Remaining 4 women can be grouped into two couples in 2!⋅2!⋅2!4!​=3 ∴ total =7C3​⋅3!⋅3=630

3 women can be selected in $^{7} C_{3}$ ways and can be paired with 3 men in 3 ! ways.
Remaining 4 women can be grouped into two couples in $\frac{4 !}{2 ! \cdot 2 ! \cdot 2 !} = 3$
$∴$ total $=^{7} C_{3} \cdot 3! \cdot 3 = 630$