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

- $C(n,r)=\large\frac{n!}{r!(n-r)!}$
- $n!=n(n-1)(n-3).....(3)(2)(1)$

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.

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