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# From a class of 25 students 10 are to be chosen for an excursion party.There are 3 students who decide that either all of them will join or none of them will join.In how many ways can the excursion party be chosen?

$\begin{array}{1 1}(A)\;817090\\(B)\;817190\\(C)\;827280\\(D)\;837320\end{array}$

Can you answer this question?

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• $C(n,r)=\large\frac{n!}{r!(n-r)!}$
There are two cases.
(a) If the 3 students join the excursion party then the number of combinations will be
$P_1=C(22,7)$
(b) If the 3 students do not join the excursion party.
Then the number of combinations
$P_2=C(22,10)$
If P is the combinations of choosing the excursion party,then
$P=P_1+P_2$=C(22,7)+C(22,10)
$\;\;\;=\large\frac{22!}{7!8!}+\frac{22!}{10!12!}$
$\;\;\;=\large\frac{22\times 21\times 20\times 19\times 18\times 17\times 16\times 15!}{7\times 6\times 5\times 4\times 3\times 2\times 1\times 15!}+\large\frac{22\times 21\times 20\times 19\times 18\times 17\times 16\times 15\times 13\times 12!}{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\times 12!}$
$\;\;\;=817190$
Hence (B) is the correct answer.
answered May 14, 2014

+1 vote