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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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18 Apr 2017, 04:54
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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
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18 Apr 2017, 18:37
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 answerWhy? Let's try with a smaller number.. 5 Choosing one out of 5 or 4 out of 5 is SAME. \(5C1=\frac{5!}{1!(51)!}=\frac{5!}{1!/4!}=5C4\).... 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 k1 gives 4851 ways.. \(99C(k1)=4851=99*49=\frac{99*98}{2}=\frac{99*98*97*96*.....*2*1}{2*97*96*..*2*1}=\frac{99!}{2!97!}\) So k1 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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Re: If Jay has 99 problems, in how many ways can he select k of them to ra
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28 Jul 2018, 07:37
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
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28 Jul 2018, 09:10
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 10k out of 10 so 10C4 = 10C(104).. Why 10C4=10!/(4!*(104)!)=10!/4!6! And 10C6 = 10!/6!(106)!=10!/6!4! Both same , So k can be 4 or 6 as both will give same value
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Re: If Jay has 99 problems, in how many ways can he select k of them to ra
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28 Jul 2018, 09:25
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
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10 Jan 2019, 06:42
Couldn't help but smile while solving this question. Almost could hear Jay's rap and the tune of the amazing song that this problem subtly refers to https://youtu.be/6uikJTnmtgwPosted from my mobile device
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Re: If Jay has 99 problems, in how many ways can he select k of them to ra
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11 Jan 2019, 08:18
Gladiator59 wrote: Almost could hear Jay's rap and the tune of the amazing song that this problem subtly refers to https://youtu.be/6uikJTnmtgwGreat choice/link, Gladiator59 !
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Re: If Jay has 99 problems, in how many ways can he select k of them to ra
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11 Jan 2019, 08:25
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.
Let me solve a "different" problem. In the end, I am sure you will understand the solution to the original problem throughly! GMATH wrote: If Jay has 7 problems, in how many ways can he select k of them to rap about?
(1) Jay can select k+1 of his problems in 35 different ways. (2) Jay can select k–1 of his problems in 21 different ways.
\(? = C\left( {7,k} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\boxed{\,? = k\,}\) \(\left( 1 \right)\,\,\,C\left( {7,k + 1} \right) = 35\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,k + 1 = 4\,\,\,{\text{or}}\,\,\,k + 1 = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = 3\,\,\,\,{\text{or}}\,\,\,\,k = 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUFF}}.\) \(\left( * \right)\,\,C\left( {7,4} \right) = C\left( {7,7  4} \right) = 35\,\,\,\,\,\,\) \(\left( 2 \right)\,\,\,C\left( {7,k  1} \right) = 21\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,k  1 = 2\,\,\,{\text{or}}\,\,\,k  1 = 5\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = 3\,\,\,\,{\text{or}}\,\,\,\,k = 6\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUFF}}{\text{.}}\) \(\left( {**} \right)\,\,C\left( {7,2} \right) = C\left( {7,7  2} \right) = 21\) \(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{ \,\left( 1 \right)\,\,\, \Rightarrow \,\,\,k = 3\,\,\,\,{\rm{or}}\,\,\,\,k = 2 \hfill \cr \,\left( 2 \right)\,\,\, \Rightarrow \,\,\,k = 3\,\,\,\,{\rm{or}}\,\,\,\,k = 6 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,k = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}{\rm{.}}\,\,\,\,\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Re: If Jay has 99 problems, in how many ways can he select k of them to ra
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11 Jan 2019, 13:48
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. One of the things that this problem can test is the concept that factorials can be unpacked into the product of n1 multiples The question asks 99Ck=? (1) This statement tells us that 99C(k+1) =3764376> 99!/((k+1)!(99(k+1))! This can be very different numbers depending on the choice of k NS (2) This statement tells us that 99C(k1)=4851  99!/(k1)!(99(k1))! = 4851. NS for the same reason as (1) (1) and (2) if we divide (1) by (2) we get {99!/((k+1)!(99(k+1)!}/{(99!/((k1)!(99(k1))!} = 3764376/4851. At first glance this seems like it might not help, however we can expand some of our factorial terms to allow us to cancel out numerators and denominators > 99!/(k+1)(k)(k1)!(98k)! * (k1)!(100k)!/99! = 3764376/4851 > (100k)!/(k(k+1))(98k)! = 3764376/4851 > (100k)(100(k+1))(100(k+2))!/(k(k+1))(98k)! = 3764376/4851 > (100k)(99k)(98k)!/((k(k+1)(98k)!) = 3764376/4851, allowing our (98k)! to cancel out. > (100k)(99k)/(k)(k+1) =3764376/4851 which is messy, but is 1V1E, so suff The answer is C



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Re: If Jay has 99 problems, in how many ways can he select k of them to ra
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12 Jan 2019, 10:29
Yes, since the combinatorics formula gives you the same answer for two values of K, then each is insufficient. Then using both you take the value in common. C.
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Re: If Jay has 99 problems, in how many ways can he select k of them to ra
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12 Jan 2019, 14:35
fskilnik wrote: 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.
Let me solve a "different" problem. In the end, I am sure you will understand the solution to the original problem throughly! GMATH wrote: If Jay has 7 problems, in how many ways can he select k of them to rap about?
(1) Jay can select k+1 of his problems in 35 different ways. (2) Jay can select k–1 of his problems in 21 different ways.
\(? = C\left( {7,k} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\boxed{\,? = k\,}\) \(\left( 1 \right)\,\,\,C\left( {7,k + 1} \right) = 35\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,k + 1 = 4\,\,\,{\text{or}}\,\,\,k + 1 = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = 3\,\,\,\,{\text{or}}\,\,\,\,k = 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUFF}}.\) \(\left( * \right)\,\,C\left( {7,4} \right) = C\left( {7,7  4} \right) = 35\,\,\,\,\,\,\) Hi Fabio, how did you reverse solve for k? I understand the equation and how the answer has two variables, I'm confused at how you actually found the values of the two variables quickly. I tried to solve as a quadratic equation, but that seems to be really inefficient? Thanks for your help!



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Re: If Jay has 99 problems, in how many ways can he select k of them to ra
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14 Jan 2019, 05:35
Rule: nCr = nC(nr) Examples: 5C2 = 5C(52) = 5C3. 10C3 = 10C(103) = 10C7. 99C5 = 99C(995) = 99C94. BTGmoderatorDC 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. Statement 2: Jay can select k1 of his problems in 4851 different ways.Since 4851 is relatively small, test small values for k1. Case 1: k1 = 2 From 99 problems, the number of ways to choose 2 = 99C2 = (99*98)/(2*1) = 4851. This works. Thus, it's possible that k1=2. Since 99C2 = 99C97, it's also possible that k1=97. k1=2 implies that k=3. k1=97 implies that k=98. Thus, Statement 1 offers two options: k=3 or k=98. INSUFFICIENT. Rule: For every DS problem, there must be at least ONE CASE that satisfies both statements. Statement 1: Jay can select k+1 of his problems in 3764376 different ways. Only two cases satisfy Statement 2: k=3 and k=98. At least one of these two cases must also satisfy Statement 1. Test k=98. Case 2: k=98, implying that k+1=99 From 99 problems, the number of ways to choose 99 = 99C99 = 1. k+1=99 does not satisfy the condition that the number of ways to choose k+1 problems is 3764376. Thus, Case 2 is not viable. Implication: Since there must be at least one case that satisfies both statements, it MUST be possible in Statement 1 that k=3, with the result that k+1=4. Since it's possible in Statement 1 that k+1=4, we know that 99C4 = 3764376. Since 99C4 = 99C95, it must also be possible in Statement 1 that k+1=95, with the result that k=94. Thus, Statement 1 offers two options: k=3 or k=94. INSUFFICIENT. Statements combined:Only the blue option above satisfies both statements, implying that k=3. SUFFICIENT.
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Re: If Jay has 99 problems, in how many ways can he select k of them to ra
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