Data Sufficiency

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.