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N, R both are positive integer and permutations P(N,R)

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N, R both are positive integer and permutations P(N,R) [#permalink] New post 16 Jan 2006, 03:15
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N, R both are positive integer and permutations P(N,R) represents N!/(N-R)! what is the minimum number of R that makes P(N,R) less than 1000?

A. 1

B. 2

C. 3

D. 4

E. 5
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 [#permalink] New post 16 Jan 2006, 06:09
Don't know if I've understand the question correctly, but one thing is clear. The less R the less P(N,R).

Assume R=0 you'll see that the result is 1
Assume R=N you'll see that the result is N!

I go for A
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 [#permalink] New post 16 Jan 2006, 09:12
Doesn't it depend on N?

If N!/(N-R)! = N*(N-1)*(N-2)*...*(N-R+1)*(N-R)!/((N-R)!
= N*(N-1)*(N-2).......*(N-R+1) ---- (1)

(1) can be less than 1000 in several cases sependong on N.

For N<32, R=1
For N>32, R=0 is the only solution.
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 [#permalink] New post 17 Jan 2006, 07:47
anyone else wanna try?! :wink:
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 [#permalink] New post 17 Jan 2006, 09:07
N! / (N-R!) = N * N-1 * N-2....*(N-R+1)

So clearly depending of values of N situation changes, So for every R we can find N which maches P(N,R)<1000 as well as P(N,R)>1000

so my pick is Min value of R=1 for N=999.

And also because 1 is smallest available in answer list that solves requirement...

Chirag
  [#permalink] 17 Jan 2006, 09:07
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N, R both are positive integer and permutations P(N,R)

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