Question

There are 10 persons named $P_1,P_2,P_3,\dots ,P_{10}$ .
Out of $10$ persons, $5$ persons are to be arranged in a line such that in each arrangement $P_1$ must occur whereas $P_4$ and $P_5$ do not occur. Find the number of such possible arrangements.

Answer 1

Given : $P_1,P_2,P_3,\dots ,P_{10}$ are $10$ persons
out of which $5$ persons are to be arranged but $P_1$ must occur and $P_4$ and $P_5$ do not occur.

As, $P_1$ is already occurring, selection to be done only for $4$ persons.

Total persons excluding $P_1,P_4$ and $P_5=7$

No. of selection $={}^7{C}_4=\frac{7!}{(7-4)!\times 4!}$

$(\because {}^n{C}_r=\frac{n!}{r!(n-r)!})$

No.of selections $=\frac{7\times 6\times 5}{3\times 2\times 1}=35$

$5$ People can be arranged in $5!$ ways

So, the number of arrangements

$=35\times 5!$

$=35\times 120$

$=4200$