Q. 25.0( 4 Votes )

# There are 10 persons named P_{1}, P_{2}, P_{3},……P_{10}. Out of 10 persons, 5 persons are to be arranged in a line such that is each arrangement P_{1} must occur whereas P_{4} and P_{5} do not occur. Find the number of such possible arrangements.

Answer :

Given that 5 persons need to be selected from 10 person P_{1}, P_{2}, P_{3},……P_{10}.

It is also told that P_{1} should be present and P_{4} and P_{5} should not be present.

It is similar to choosing 4 persons from remaining 7 persons as P_{1} is selected and P_{4} and P_{5} are already removed.

Let us first find the no. of ways to choose persons and assume it to be N_{1}.

⇒ N_{1} = Selecting 4 persons from remaining 7 persons

⇒ N_{1} = ^{7}C_{4}

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N_{1} = 35

Now we need to arrange the chosen 5 people. Since 1 person differs from other.

The arrangement is similar to that of arranging n people in n places which are n! Ways to arrange. So, the persons can be arranged in 5! Ways.

Let us assume the total possible arrangements be N.

⇒ N = N_{1} × 5!

⇒ N = 35 × 120

⇒ N = 4200

∴ The total no. of possible arrangement can be done is 4200.

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