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 -
Text Solution
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For 1 ≤ i ≤ 4, let x i ( ≥ 3) be the number of blanks between i th and (i + 1) th letters. Then,
x 1 + x 2 + x 3 + x 4 = 15 …..(1)
The number of solutions of (1)
= coefficient of t 15 in (t 3 + t 4 +….) 4
= coefficient of t 3 in (1 – t) –4
= coefficient of t 3 in [1 + 4 C 1 + 5 C 2 t 2 + 6 C 3 t 3 + …..]
= 6 C 3 = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence is correct answer.
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