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If Jay has 99 problems, in how many ways can he select k of them to ra

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If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

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New post 18 Apr 2017, 03:54
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Question Stats:

44% (01:45) correct 56% (01:24) wrong based on 294 sessions

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If Jay has 99 problems, in how many ways can he select k of them to rap about?

(1) Jay can select k+1 of his problems in 3764376 different ways.
(2) Jay can select k–1 of his problems in 4851 different ways.

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If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

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New post 18 Apr 2017, 17:37
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Bunuel wrote:
If Jay has 99 problems, in how many ways can he select k of them to rap about?

(1) Jay can select k+1 of his problems in 3764376 different ways.
(2) Jay can select k–1 of his problems in 4851 different ways.



Hi,

We do not and we should not get overwhelmed by the huge numbers here..
the main point is that we require two figures here to get on to value of k, and thus C should be the answer

Why?
Let's try with a smaller number.. 5
Choosing one out of 5 or 4 out of 5 is SAME.
\(5C1=\frac{5!}{1!(5-1)!}=\frac{5!}{1!/4!}=5C4\)....
So it can be 1 or 4, and therefore we require TWO relations of k to get to the value of k...

Let's see the statements..
First let's check the smaller number
II. Choosing k-1 gives 4851 ways..
\(99C(k-1)=4851=99*49=\frac{99*98}{2}=\frac{99*98*97*96*.....*2*1}{2*97*96*..*2*1}=\frac{99!}{2!97!}\)
So k-1 can be 2 or 97, thus k can be 2+1=3 OR 97+1=98..
Insuff alone

I. Choosing k+1is 3764376...
Insufficient

Combined..
If k is 98, k+1 is 99... BUT choosing 99 out of 99 is 1 so eliminate
Therefore k+1 must be 3+1=4 and k MUST be 3
Sufficient

C
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

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New post 28 Jul 2018, 06:37
Hi Chetan

Could you please elaborate on the first statement. How did you determine its insufficiency?
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If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

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New post 28 Jul 2018, 08:10
Rw27 wrote:
Hi Chetan

Could you please elaborate on the first statement. How did you determine its insufficiency?


Hi when you are looking for combinations..
Choosing k out of say 10 will be same as finding 10-k out of 10 so 10C4 = 10C(10-4)..
Why 10C4=10!/(4!*(10-4)!)=10!/4!6!
And 10C6 = 10!/6!(10-6)!=10!/6!4!
Both same ,

So k can be 4 or 6 as both will give same value
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

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New post 28 Jul 2018, 08:25
Ohok, got it. Thanks
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Re: If Jay has 99 problems, in how many ways can he select k of them to ra &nbs [#permalink] 28 Jul 2018, 08:25
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