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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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There are 10 persons named $P_1,P_2,P_3.....P_{10}$.Out of 10 persons,5 person are to be arranged in a line such that in each arrangement $P_1$ must occur where as $P_4$ and $P_5$ do not occur.Find the number of such possible arrangement.

$\begin{array}{1 1}(A)\;4200\\(B)\;4300\\(C)\;4400\\(D)\;4500\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
  • $n!=n(n-1)(n-3).....(3)(2)(1)$
10 person named $P_1,P_2,P_3......P_{10}$
5 persons are to be arranged in a line
No of ways in which $P_1$ must occur =$5!$
$\Rightarrow 5\times 4\times 3\times 2\times 1$
$\Rightarrow 120$ways
No of ways of the remaining persons are $=7C_4$
$\Rightarrow \large\frac{7!}{4!3!}$
$\Rightarrow \large\frac{7\times 6\times 5\times 4!}{4!\times 3\times 2\times 1}$
$\Rightarrow 35$
Total no of ways =$120\times 35$
$\Rightarrow 4200$ ways
Hence (A) is the correct answer.
answered May 15, 2014 by sreemathi.v

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