Question

Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is –

• A. 1200
• B. 1800
• C. 2400
• D. 3000

Answer 1

Answer: (C)

2400

Answer 2

When 1 £ i £ 4, let x_i (³ 3) be the number of blanks between i^th and (i + 1)^th letters.

Thus x₁ + x₂ + x₃ + x₄ = 13

No. of solution of (1)

= Coefficient of x¹⁵ in (x³ + x⁴ + ….)⁴

= Coefficient of x³ in (1 – x)^–4

= Coefficient of x³ in (1 + ⁴C₁ + ⁵C₂x² + ⁶C₃ x³ + …)

= ⁶C₃ = 20

But 5 letters can be permuted in 5! i.e. 120 ways.

Hence the reqd. number of arrangements

= (20) (120) = 2400.