What is the total number of permutations of n different things taken not more than p times, when each thing may be repeated any number of times? A. n^p +1 B. n(n^p-1) C.n^p (n+1)/(n-1) D. n(n^p - 1)/ (n-1) E. n^p (n+1)

is the answer B? n different things- can be repeated in n different ways but only p-1 times therefore n(n^p-1)

The best way to solve such questions is to substitute values for n and p.

The answer will be n+n^2+n^3+....+n^p = n(n^p-1)/n-1 gabriel wrote:The best way to solve such questions is to substitute values for n and p. Yes, this is definitely true. Anyway, let me try to explain the concept here. I guess every aspirant is aware of the following basic concepts. Total number of permutations of r things from n distinct things when repetition is not allowed is nPr Total number of permutations of r things from n distinct things when repetition is allowed is n^r Now the given question is : we have arrange at least 1 thing and at most p things from n things and it is clearly mentioned that repetition is allowed is allowed. 1 thing can be arranged from n things in n^1 ways 2 things can be arranged from n things in n^2 ways. ............................................ ......................................... p things can be arranged from n things in n^p ways. So the answer is n+n^2+n^3+......+n^p This is a G.P with common ratio n (n>1) So the sum to p terms of this G.P is n(n^p-1)/n-1 Hence D

n items taken 1 time = n n items taken 2 times where items can repeat is possible = n.n = n^2 so total permutation = n + n^2 + n^3 + .... n(n^p-1)/(n-1)