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

a) 1200
b) 1800
c) 2400
d) 3000

Answer 1

 The min no of blanks between every two letter are=3 and as the total no of blanks are 15 and hence the max no of blanks that may contain there are=6. Hence using Binomial and GP series expansion we get the no of ways to place the 15 gaps are = 6C3= 20 and the no of ways of arrangeing 5 different letters are= 5!=120
Hence the no of different arrangements are=(120×20)=2400(ans)

Answer 2

Answer:

c) 2400

Step-by-step explanation:

There are 5 different symbols so those symbols can be arranged in 5! = 120 ways.

Now we have to place blank spaces between 5 different symbols with at least 3 spaces between each symbol so

So, let a,b,c,d,e be the 5 distinct symbols.

 a _ _ _ b _ _ _ c _ _ _ d _ _ _ e

As we can see that if we put 3 spaces between each symbol then 4 x 3 = 12 symbols will be places as it is. Now we have to place remaining 15 - 12 = 3 blanks in 4 places that is between different symbols

So to place r identical objects in n different boxes formula is (n+r-1)C(r)

(4+3-1)C(3) = 6C3 = 20.

So total number of ways are 120 x 20 = 2400.

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