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.
OA C
Source: Veritas Prep
If Jay has 99 problems, in how many ways can he select k of
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We have to find out the value of 99Ck. If we know the unique value of k, we get the answer.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.
OA C
Source: Veritas Prep
Question: What's the value of k?
Let's take each statement one by one.
(1) Jay can select k+1 of his problems in 3764376 different ways.
=> 99C(k+1) = 3764376
Though this is a big number, it will give us the value of k. However, would the value of k unique? The answer is No.
It is because since nCr = nC(n-r)
Let's understand this an easier example.
We know that
> 5C3 = (5.4.3)/(1.2.3) = 10; however,
> 5C2 = (5.4)/(1.2) = 10. So, the value of r is not unique.
This means that there would be two values of k. No unique value of k. Insufficient.
(2) Jay can select k-1 of his problems in 4851 different ways.
As with Statement (1), the same is with Statement 2. Insufficient.
(1) and (2) together
Combining the two statements will give us a common value of k from the two statements. Sufficient.
The correct answer: C
Hope this helps!
-Jay
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Let me solve a "different" problem. In the end, I am sure you will understand the solution to the original problem throughly!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.
Source: Veritas Prep
$$? = C\left( {7,k} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\boxed{\,? = k\,}$$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.
$$\left( 1 \right)\,\,\,C\left( {7,k + 1} \right) = 35\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,k + 1 = 4\,\,\,{\text{or}}\,\,\,k + 1 = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = 3\,\,\,\,{\text{or}}\,\,\,\,k = 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUFF}}.$$
$$\left( * \right)\,\,C\left( {7,4} \right) = C\left( {7,7 - 4} \right) = 35\,\,\,\,\,\,$$
$$\left( 2 \right)\,\,\,C\left( {7,k - 1} \right) = 21\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,k - 1 = 2\,\,\,{\text{or}}\,\,\,k - 1 = 5\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = 3\,\,\,\,{\text{or}}\,\,\,\,k = 6\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUFF}}{\text{.}}$$
$$\left( {**} \right)\,\,C\left( {7,2} \right) = C\left( {7,7 - 2} \right) = 21$$
$$\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\,\left( 1 \right)\,\,\, \Rightarrow \,\,\,k = 3\,\,\,\,{\rm{or}}\,\,\,\,k = 2 \hfill \cr
\,\left( 2 \right)\,\,\, \Rightarrow \,\,\,k = 3\,\,\,\,{\rm{or}}\,\,\,\,k = 6 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,k = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}{\rm{.}}\,\,\,\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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