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 is each arrangement P₁ must occur whereas P₄ and P₅ do not occur. Find the number of such possible arrangements.

Answer 1

Given as

The total persons = 10

The number of persons to be selected = 5 from 10 persons (P₁, P₂, P₃ … P₁₀)

It is also told that P₁ should be present and P₄ and P₅ should not be present.

Now, we have to choose 4 persons from remaining 7 persons as P₁ is selected and P₄ and P₅ are already removed.

The number of ways = Selecting 4 persons from remaining 7 persons

= ⁷C₄

By using the formula,

ⁿC_r = n!/r!(n – r)!

⁷C₄ = 7! / 4!(7 – 4)!

= 7! / (4! 3!)

= [7 × 6 × 5 × 4!] / (4! 3!)

= [7 × 6 × 5] / (3 × 2 × 1)

= 7 × 5

= 35

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

35 × 5! = 35 × (5 × 4 × 3 × 2 × 1)

= 4200

Hence, the total no. of possible arrangement can be done is 4200.