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# For any integer P greater than 1, P! denotes the product of all the in

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ISB, NUS, NTU Moderator
Joined: 11 Aug 2016
Posts: 376
For any integer P greater than 1, P! denotes the product of all the in  [#permalink]

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05 Sep 2018, 12:15
3
00:00

Difficulty:

65% (hard)

Question Stats:

53% (02:06) correct 47% (01:33) wrong based on 20 sessions

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For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. In how many ways can 9! be expressed as a product of 2 positive integers?

A. 36
B. 72
C. 79
D. 80
E. 160

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Joined: 01 Oct 2017
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Re: For any integer P greater than 1, P! denotes the product of all the in  [#permalink]

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05 Sep 2018, 12:36
1
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. In how many ways can 9! be expressed as a product of 2 positive integers?

A. 36
B. 72
C. 79
D. 80
E. 160

9!=1*2*3*4*5*6*7*8*9
Writing in prime factorization form,
$$9!=2^7*3^4*5*7$$
No of distinct factors of 9! is (7+1)(4+1)(1+1)(1+1)=8*5*2*2=160
We need to find out the pair of positive integers whose product is 9!.
So, No of ways we can express 9! as a product of 2 +ve integers=$$\frac{160}{2}$$=80

Ans. (D)
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PKN

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Re: For any integer P greater than 1, P! denotes the product of all the in   [#permalink] 05 Sep 2018, 12:36
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# For any integer P greater than 1, P! denotes the product of all the in

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