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A man can invite exactly $3$ girls from $5$ girls $=\;^{5}C_{3} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!}=10$ ways

Now, boys can be invited $0,1,2,3, \text{(or)}\;4$

Number of ways boys can be invited $ = \;^{4}C_{0} + \;^{4}C_{1} + \;^{4}C_{2} + \;^{4}C_{3} + \;^{4}C_{4} = 1 + 4 + 6 + 4 +1 = 16$ ways

$\therefore$ The total number of ways $ = 10 \times 16 = 160$ ways.

Correct Answer $:160$

$\textbf{PS:}$

- $\;^{n}C_{0} + \;^{n}C_{1} + \;^{n}C_{2} + \dots + \;^{n}C_{n} = 2^{n}$
- The number of ways to pick $‘r\text{’}$ unordered element from an $‘n\text{’}$ element set is $\;^{n}C_{r} = \frac{n!}{(n-r)!r!}$