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

Question

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.

chetan2u · Accepted Answer

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

Gladiator59 · Answer

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://www.youtube.com/watch?v=6uikJTnmtgw

Rw27 · Answer

Hi Chetan

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

chetan2u · Answer

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

fskilnik · Answer

Let me solve a "different" problem. In the end, I am sure you will understand the solution to the original problem throughly!

? = C\left( {7,k} ight)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\boxed{\,? = k\,}

\left( 1 ight)\,\,\,C\left( {7,k + 1} ight) = 35\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,k + 1 = 4\,\,\,{	ext{or}}\,\,\,k + 1 = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = 3\,\,\,\,{	ext{or}}\,\,\,\,k = 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{	ext{INSUFF}}.

\left( * ight)\,\,C\left( {7,4} ight) = C\left( {7,7 - 4} ight) = 35\,\,\,\,\,\,

\left( 2 ight)\,\,\,C\left( {7,k - 1} ight) = 21\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} ight)} \,\,\,\,\,k - 1 = 2\,\,\,{	ext{or}}\,\,\,k - 1 = 5\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = 3\,\,\,\,{	ext{or}}\,\,\,\,k = 6\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{	ext{INSUFF}}{	ext{.}}

\left( {**} ight)\,\,C\left( {7,2} ight) = C\left( {7,7 - 2} ight) = 21

\left( {1 + 2} ight)\,\,\,\left\{ \matrix{
 \,\left( 1 ight)\,\,\, \Rightarrow \,\,\,k = 3\,\,\,\,{m{or}}\,\,\,\,k = 2 \hfill \cr 
 \,\left( 2 ight)\,\,\, \Rightarrow \,\,\,k = 3\,\,\,\,{m{or}}\,\,\,\,k = 6 \hfill \cr} ight.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,k = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{m{SUFF}}{m{.}}\,\,\,\,

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

GMATGuruNY · Answer

Rule:
nCr = nC(n-r)

Examples:
5C2 = 5C(5-2) = 5C3.
10C3 = 10C(10-3) = 10C7.
99C5 = 99C(99-5) = 99C94.

Statement 2: Jay can select k-1 of his problems in 4851 different ways. 
Since 4851 is relatively small, test small values for k-1.
Case 1: k-1 = 2
From 99 problems, the number of ways to choose 2 = 99C2 = (99*98)/(2*1) = 4851.
This works.
Thus, it's possible that k-1=2.
Since 99C2 = 99C97, it's also possible that k-1=97.
k-1=2 implies that k=3.
k-1=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.

C

ueh55406 · Answer

still waiting for that one clear explaination:

I got D. both are suff.

from st 1 > 99 !/ (99-k+1)!* k+1! = 3764376

from st 2> 99!/(99-(k-1)!*K-1!= 4851.

using both the stmts we can find a unique value for K? and we should have D as the answer. 
would appreciate if someone could help out here.

GMATGuruNY · Answer

As shown in my earlier post:
Statement 1 is satisfied by two values: k=3 and k=94.
Statement 2 is also satisfied by two values: k=3 and k=98.
Thus, each statement on its own is INSUFFICIENT.
Only k=3 satisfies BOTH statements.
Thus, the two statements COMBINED are sufficient to determine that k=3.

C

sarthak2036 · Answer

But how do we know for sure that one of the two k values in each statement will overlap? Can it not be the case that both statements give four different values of k?

Please help!

chetan2u · Answer

There is one exact value of k as per the main statement.

Statement I will give us 2 values, one this exact value of k and some other value, total-exact value. 
Statement II would also do the same thing. One exact value and some other value. 
The common in two statement will be this EXACT value we are looking for. 
You cannot have k having different values in the different statements in actual GMAT.

It is similar to:
What is x?
1) some quadratic equation say x^2-7x+12=0
2) some quadratic equation say x^2-x-6=0

Now each of the statements will give two values of x, and x has to be one of them. 
So the common value will give you the answer.

But say one statement was x^2-6x+9=0....(x-3)^2=0. Both the values of x are 3. So sufficient on its own. 
If the statement II in this case too gives you two values of x. One would surely be 3. Otherwise the two statements do not match as they do not give same value of x. 
Few unofficial questions do have different values in each statement, a situation that is wrong as per actual GMAT.

einstein801 · Answer

Hi chetan, how would you realistically break up 4851 to  99*49?

chetan2u · Answer

unicornilove 
Since we are talking of nC(k-1) being equal to 4851, it has to be a multiple of n and n is 99 here. 
Also, the divisibility rule of 9 and 11 will tell you that 4851 is a multiple of both 9 and 11. 
9: 4+8+5+1 is a multiple of 9.
11: (4+5)-(8+1) is a multiple of 11

Now once we know 4851 will be a multiple of 99, we can home on to the second factor. 
4851=99*ab, where b has to be 9, that is 4851*a9
4851 is close to 5000 or 100*50, so 99*49 is the most likely answer between possible 99*39, 99*49 or 99*59.

Generally number properties should get you closer to the answer faster.

samarpan.g28 · Answer

If we take the statement 1, 99C(k+1)= 99!/(k+1)!*(98-k)! =3764376....(i) Insufficient.
From the statement 2, 99C(k-1)= 99!/(k-1)!*(100-k)! =4851....(ii) Insufficient.

Divide (i) by (ii),  (k-1)!*(100-k)!/(k+1)!*(98-k)! = 3764376/4851 =776
or,  (100-k)*(99-k)/k*(k+1) =776
or, (100-k)*(99-k)=776*k*(k+1) - we can find the value of k from here. Option (C) is correct.

arunbhati · Answer

Hi Bunuel,

How we will realise 4851 is nothing but 99*49 in exam under time pressure

any other way to tackle such type of problem

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

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GMAT Data Sufficiency (DS) Questions

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

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