There are 10 persons named P1, P2, P3,……P10. Out of 10 persons, 5 persons are to be arranged in a line such that is each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possible arrangements.
Given that 5 persons need to be selected from 10 person P1, P2, P3,……P10.
It is also told that P1 should be present and P4 and P5 should not be present.
It is similar to choosing 4 persons from remaining 7 persons as P1 is selected and P4 and P5 are already removed.
Let us first find the no. of ways to choose persons and assume it to be N1.
⇒ N1 = Selecting 4 persons from remaining 7 persons
⇒ N1 = 7C4
We know that 
,
And also n! = (n)(n – 1)......2.1
⇒ 
⇒ 
⇒ 
⇒ N1 = 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 = N1 × 5!
⇒ N = 35 × 120
⇒ N = 4200
∴ The total no. of possible arrangement can be done is 4200.