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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