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five distinct letters are to be transmitted through a communication channel.a total number of 15 blanks is to be inserted between the 2 letters with atleast 3 between every 2.the number of ways in which this can be done is? please reply soon.............
Dear Sri Valli,
Ans:- 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)
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All the best Sri Valli!!!
Regards,
Askiitians ExpertsSoumyajit Das IIT Kharagpur
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