Question

Five distinct letters are supposed to be transmitted through a communication channel. A total number of 15 blanks is supposed to be inserted between the first and the last letter with at least three between every two. The number of ways in which this can be done.
A. 1200
B. 1800
C. 2400
D. 3000

Answer 1

We will first find the number of ways in which the 5 letters can be arranged. Then, arrange 3 blanks between every two numbers, and there will only 3 blanks left that we have to arrange. Let the number of blanks that are to be arranged between two numbers be ${a_1},{a_2},{a_3},{a_4}$, and equate it to three. Then, find the number of integral solutions to the equation.

Complete step by step Answer:

We are given that 5 letters are to be arranged. Hence, there are $n$ distinct numbers, then there are $n!$ ways to arrange it.
Hence, there are 5! ways to arrange 5 letters.
But, there is a total number of 15 blanks are supposed to be inserted between the first and the last letter with at least three between every two.
There are gaps in which blanks can be arranged.
If we have to place 3 blanks between every 2 letters.
Then the number of blanks that have to place between the letters are $3 \times 4 = 12$ blanks
That is, there are 3 blanks left that need to be arranged.
Let the number of blanks inserted between first numbers from 3 blanks is ${a_1}$, between second and third number be ${a_2}$, between third and fourth number be ${a_3}$, and between the fourth and fifth number is ${a_4}$
Also, the total of these blanks has to be 3
That is ${a_1} + {a_2} + {a_3} + {a_4} = 3$
We have to find the number of integral solutions of the above equation.
$^{3 + 4 - 1}{C_{4 - 1}}{ = ^6}{C_3}$
The formula of $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Then, $\dfrac{{6!}}{{2!4!}} = \dfrac{{6.5.4!}}{{2.1.4!}} = 20$
The total number of ways is $5! \times 20 = 5.4.3.2.1.20 = 2400$
Hence, the option (c) is correct.

Note: Many students forget the possibility that the numbers will also arrange among themselves. If there are $n$ distinct numbers, then there are $n!$ ways to arrange it. If there are 5 letters, the number of ways they can be arranged is 5! We have to multiply this with the number of ways to arrange blanks for the final answer.