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 -
- 1200
- 1800
- 2400
- 3000
The correct answer is: 2400
For 1 i 4, let xi ( 3) be the number of blanks between ith and (i + 1)th letters. Then,
x1 + x2 + x3 + x4 = 15 …..(1)
The number of solutions of (1)
= coefficient of t15 in (t3 + t4 +….)4
= coefficient of t3 in (1 – t)–4
= coefficient of t3 in [1 + 4C1 + 5C2 t2 + 6C3 t3 + …..]
= 6C3 = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence
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