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

Solution :
$$10 < _{7w}^{3m}3$$   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 $$\frac{{4!}}{{\left( {2! \cdot 2! \cdot 2!} \right)}} = 3.$$
Therefore, total $$ = {\,^7}{C_3} \cdot 3! \cdot 3 = 630.$$