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

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SVP
Joined: 24 Sep 2005
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N, R both are positive integer and permutations P(N,R) [#permalink]  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
Director
Joined: 17 Dec 2005
Posts: 548
Location: Germany
Followers: 1

Kudos [?]: 16 [0], given: 0

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
VP
Joined: 21 Sep 2003
Posts: 1065
Location: USA
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Kudos [?]: 40 [0], given: 0

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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SVP
Joined: 24 Sep 2005
Posts: 1891
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anyone else wanna try?!
Senior Manager
Joined: 05 Jan 2006
Posts: 382
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Kudos [?]: 38 [0], given: 0

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
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