There are 5 women and 3 men waiting on standby for a flight to New York.

We need to select 3 of the 8 people. Since the order of the people is not important (as a different order results in the same people being selected), we need to use a combination.
Definition combination (order is not important);$nCr=(nr)=\frac{n!}{r!(n-r)!}$ with $n!=n(n-1)\cdot \dots \cdot 2\cdot 1$. # of possible outcomes=$s{C}_3=\frac{8!}{3!(8-3)!}=\frac{8!}{3!5!}=56$
We need to select 2 of the 5 women and 1 of the 3 men to obtain $X=2$. # of favorable outcomes=$(5C_2\cdot 3)C_1=(\frac{5!}{2!3!})\cdot \frac{3!}{1!2!}=10\cdot 3=30$
The probability is the number of favorable outcomes divided by the number of possible outcomes: $Pr(X=2)=$$\frac{\text{\# of favorable outcomes}}{\text{ \# of possible outcomes}}$ $=\frac{30}{56}=\frac{15}{28}$ $0.5357=53.57\mathrm{\%}$

We need to select 3 of the 8 people. Since the order of the people is not important (as a different order results in the same people being selected), we need to use a combination.
Definition combination (order is not important);${\displaystyle nCr=\left(nr\right)=n\frac{!}{r!(n-r)!}}$ with ${\displaystyle n\ne n(n-1)\cdot \dots 2\cdot 1}$. # of possible outcomes=${\displaystyle sC3=\frac{8!}{3!(8-3)!}=\frac{8!}{3}!5\ne 56}$
We need to select 2 of the 5 women and 1 of the 3 men to obtain X=2. # of favorable outcomes=${\displaystyle (5C_2\cdot 3)C_1=(\frac{5!}{2}!3!)\cdot \frac{3!}{1}!2\ne 10\cdot 3=30}$
The probability is the number of favorable outcomes divided by the number of possible outcomes: $\frac{\text{\# of favorable outcomes}}{\text{ \# of possible outcomes}}$ $=30/56=15/28=0.5357=53.57$