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taken not more than p times

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by sanju09 » Fri Feb 20, 2009 3:18 am
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)
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by kanha81 » Fri Feb 20, 2009 2:24 pm
is the answer B?

n different things- can be repeated in n different ways but only p-1 times
therefore n(n^p-1)
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by gabriel » Fri Feb 20, 2009 5:10 pm
The best way to solve such questions is to substitute values for n and p.
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by sureshbala » Fri Feb 20, 2009 8:37 pm
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
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by Bidisha_800 » Fri Feb 20, 2009 11:50 pm
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)
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