QUESTION 178
The value of $5! - 3!$ is
(A) 108
(B) 112
(C) 114
(D) 116
(E) 120

Find the natural number $n$ that satisfies the equation $2 \times {}_{n}\mathrm{C}_{3} = 3 \times {}_{n}\mathrm{P}_{2}$. [3 points]

Find the value of ${ } _ { 5 } \mathrm { P } _ { 2 } + { } _ { 5 } \mathrm { C } _ { 2 }$. [3 points]

Find the value of ${ } _ { 6 } \mathrm { P } _ { 2 } - { } _ { 6 } \mathrm { C } _ { 2 }$. [3 points]

Find the value of ${}_{6}\mathrm{P}_{2} - {}_{6}\mathrm{C}_{2}$. [3 points]

Find the value of ${ } _ { 7 } \mathrm { P } _ { 2 } + { } _ { 7 } \mathrm { C } _ { 2 }$. [3 points]

The value of $\left( 2 \cdot { } ^ { 1 } P _ { 0 } - 3 \cdot { } ^ { 2 } P _ { 1 } + 4 \cdot { } ^ { 3 } P _ { 2 } - \right.$ up to $51 ^ { \text {th} }$ term $) + ( 1 ! - 2 ! + 3 ! -$ up to $51 ^ { \text {th} }$ term) is equal to
(1) $1 - 51 ( 51 )$ !
(2) $1 + ( 51 )$ !
(3) $1 + ( 52 )$ !
(4) 1

If ${ } ^ { n } P _ { r } = { } ^ { n } P _ { r + 1 }$ and ${ } ^ { n } C _ { r } = { } ^ { n } C _ { r - 1 }$, then the value of $r$ is equal to:
(1) 1
(2) 4
(3) 2
(4) 3

If ${}^{2n+1}P_{n-1} : {}^{2n-1}P_n = 11 : 21$, then $n^2 + n + 15$ is equal to $\_\_\_\_$.

For two-digit positive integers a and b
$$\frac { a ! } { b ! } = 132$$
Given this, what is the sum $\mathbf { a } + \mathbf { b }$?
A) 22
B) 23
C) 24
D) 25
E) 26

$$( n + 2 ) ! - ( n + 1 ) ! - n ! = 2 ^ { 3 } \cdot 3 \cdot 5 ^ { 2 } \cdot 7$$
Given this, what is $n$?
A) 5
B) 6
C) 7
D) 8
E) 9

$$\frac { ( 10 ! ) ^ { 2 } - ( 9 ! ) ^ { 2 } } { 11 ! - 10 ! - 9 ! }$$
Which of the following is this operation equal to?
A) $8 !$
B) $9 !$
C) $10 !$
D) $8 \cdot 8 !$
E) $8 \cdot 9 !$

$$\frac { ( n + 1 ) ! + ( n - 1 ) ! } { n ^ { 3 } - 1 } = 24$$
Given this equality, what is n?
A) 3
B) 5
C) 6
D) 8
E) 9

$\frac { 6 ! + 7 ! } { ( 4 ! ) ^ { 2 } }$\ What is the result of this operation?\ A) 10\ B) 12\ C) 15\ D) 18\ E) 20