Question

Q12. Mr. Jones owns 4 pairs of pants, 7 shirts, and 3 sweaters. In how many ways may
he choose 2 of the pairs of pants, 3 of the shirts, and 1 of the sweaters to pack for a
trip?​

Answer 1

Answer: 630 ways

Step-by-step explanation:

He can choose 2 pairs of pants in (42)=4!2!(4−2)!=6 ways, 3 of the shirts in (73)=35 ways and one of the sweater in 3 ways, complexify he can pack in 6∗35∗3=630 ways

Answer 2

Concept Introduction:-

Branch of mathematics evaluates whether an event is likely to occur or whether a proposition is likely to be true using numerical descriptions.

Given Information:-

We have been given that Mr. Jones owns $4$ pairs of pants, $7$ shirts, and $3$ sweaters.

To Find:-

We have to find that in how many ways may he choose $2$ of the pairs of pants, $3$ of the shirts, and $1$ of the sweaters to pack for a trip.

Solution:-

According to the problem

Mr. Jones can choose $2$ pairs of pants in $\left(\begin{array}{l}4 \\ 2\end{array}\right)=\frac{4 !}{2 !(4-2) !}=6$ ways, $3$ of the shirts in $\left(\begin{array}{l}7 \\ 3\end{array}\right)=35$ ways and one of the sweather in $3$ ways, complexively he can pack in $6 * 35 * 3=630$ ways

Final Answer:-

Mr. Jones can pack in $630$ ways to pack for a trip.