Question

Assertion :In a zoo, $m$ distinct animals of a circus have to be placed is $n$ cages, one in each cage. If $n$ cages $(n<m)$ are too small to accomodate $p(n<p<m),$ then the number of ways of putting the animals into cages is $\left(\begin{array}{c}m-n\\ p\end{array}\right)\left(\begin{array}{c}m-p\\ m-p\end{array}\right)$ Reason: If a work can be done by $R$ ways and another work can be done by $S$ ways then the number of ways in,which both the work can be done at a time is $R+S$.

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A),
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A)is true but (R) is false,
• D. (A)is false but (R) is true.

Answer 1

The correct option is C (A)is true but (R) is false,

Reason is false. The correct version should be:
If a work can be done by $R$ ways and another work can be done by $S$ ways, then the number of ways in,which both the work can be done at a time is $R\times S$
Assertion:
Since $n$ are too small, $m-n$ are of correct size. Thus, $p$ animals can be placed in the $m-n$ cages in ${}^{m-n}{P}_p$ ways.
There are $m-p$ cages left for the remaining $m-p$ animals which can be distributed in ${}^{m-p}{P}_{m-p}$ ways.
Thus, the answer is ${}^{m-n}{P}_p{\times }^{m-p}{P}_{m-p}$ ways.
So, the Assertion is true.
Hence (c) is correct.