Question

Number of ways in which $16$ players can be divided into four equal groups, is $K{\prod }_{r=i}^8(2r-1)$, where $K$ equals

• A. $\frac{35}{27}$
• B. $\frac{35}{24}$
• C. $\frac{35}{52}$
• D. $\frac{35}{6}$

Answer 1

The correct option is A $\frac{35}{27}$

Number of ways (m x n ) things can be divided equally into 'n' groups [with each group 'm' things]

$=\frac{\left(mn\right)!}{(m!{)}^x.n!}$

$\frac{16!}{(4!{)}^4.4!}=K[1\times 3\times 5\times 7\times 9\times 11\times 13\times 15]$

$K=\frac{16\times 14\times 12\times 10\times 8\times 6\times 4\times 2}{(24{)}^5}=\frac{35}{27}$