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

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

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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.

OA C

Source: Veritas Prep

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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
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.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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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
Note that unlike Permutation, in case of the combination, nCr = nC(n-r). Thus, we do not always get the unique value of r.

Let's take an example.

• Number of ways of choosing 3 things out of 4 = 4C3 = (4.3.2)/(1.2.3) = 4; and
• Number of ways of choosing 1 thing out of 4 = 4C1 = 4

For the same number of value, 4, we get two different values of r (3 and 1).

Let's take each statement one by one.

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

Thus, 99C(k+1) = 3764376. It is obvious that we will get two values of k. No unique answer. Insufficient.

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

Thus, 99C(k-1) = 4851. It is obvious that we will get two values of k. No unique answer. Insufficient.

(1) and (2) together

Since 99C(k+1) ≠ 99C(k-1), we will get the unique value of k. Sufficient.

The correct answer: C

Hope this helps!

-Jay
_________________
Manhattan Review

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Rule:
nCr = nC(n-r)

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

The correct answer is C.

_________________
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GMATGuruNY@gmail.com

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