In how many ways can one select a cricket team of eleven from $17$ players in which only $5$ players can bowl if each cricket team of $11$ must include exactly $4$ bowlers?
Out of $17$ players, $5$ players are bowlers.
A cricket team of $11$ players is to be selected in such a way that there are exactly $4$ bowlers.
$4$ bowlers can be selected in $^{5} C_{4}$ ways and the remaining $7$ players can be selected out of the $12$ players in $^{12} C_{7}$ ways.
Thus, by multiplication principle, required number of ways of selecting cricket team
$=\,^{5} C_{4} \times \,^{12} C_{7}=\frac{5 !}{4 ! 1 !} \times \frac{12 !}{7 ! 5 !}=5 \times \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}=3960$
Let $S=\{1,2,3,5,7,10,11\}$. The number of nonempty subsets of $S$ that have the sum of all elements a multiple of $3$ , is $........$
A committee of $12$ is to be formed from $9$ women and $8$ men in which at least $5$ women have to be included in a committee. Then the number of committees in which the women are in majority and men are in majority are respectively
If $\alpha { = ^m}{C_2}$, then $^\alpha {C_2}$is equal to
If $^{2017}C_0 + ^{2017}C_1 + ^{2017}C_2+......+ ^{2017}C_{1008} = \lambda ^2 (\lambda > 0),$ then remainder when $\lambda $ is divided by $33$ is-
For non-negative integers $s$ and $r$, let
$\binom{s}{r}=\left\{\begin{array}{ll}\frac{s!}{r!(s-r)!} & \text { if } r \leq s \\ 0 & \text { if } r>s\end{array}\right.$
For positive integers $m$ and $n$, let
$(m, n) \sum_{ p =0}^{ m + n } \frac{ f ( m , n , p )}{\binom{ n + p }{ p }}$
where for any nonnegative integer $p$,
$f(m, n, p)=\sum_{i=0}^{ p }\binom{m}{i}\binom{n+i}{p}\binom{p+n}{p-i}$
Then which of the following statements is/are $TRUE$?
$(A)$ $(m, n)=g(n, m)$ for all positive integers $m, n$
$(B)$ $(m, n+1)=g(m+1, n)$ for all positive integers $m, n$
$(C)$ $(2 m, 2 n)=2 g(m, n)$ for all positive integers $m, n$
$(D)$ $(2 m, 2 n)=(g(m, n))^2$ for all positive integers $m, n$