There are 10 persons named P1,P2,P3, … P10.

Question:

There are 10 persons named P1,P2,P3, … P10. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possible arrangements. [Hint: Required number of arrangement =7C4× 5!]

Solution:

We know that,

nCr

$=\frac{n !}{r !(n-r) !}$

According to the question,

There are 10 person named P1, P2, P3, … P10.

Number of ways in which P1 can be arranged =5! =120

Number of ways in which others can be arranged,

7C4 = 7!/(4!3!) = 35

Therefore, the required number of arrangement =C47 × 5

=35 x 120

 

=4200

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