سوال ۱۱۶ کنکور ارشد علوم کامپیوتر ۱۳۹۷

Split the universe $A=\{1,2,\ldots,15\}$ into odd and even elements.

• Odd numbers: $\{1,3,5,7,9,11,13,15\}$ — there are $8$ of them.
• Even numbers: $\{2,4,6,8,10,12,14\}$ — there are $7$ of them.

A valid subset must contain exactly three odd numbers and may contain any combination of even numbers.

Choosing the odd part. Pick exactly $3$ of the $8$ odd numbers: $\binom{8}{3}.$

Choosing the even part. Each of the $7$ even numbers is independently included or excluded, with no restriction: $2^{7}.$

By the multiplication principle the two independent choices combine: $\binom{8}{3}\cdot 2^{7}.$

This matches option 1. (Numerically $\binom{8}{3}=56$, so the count is $56\cdot 128 = 7168$.)

Exam trick: "exactly $k$ of one type, anything goes for the rest" factors as $\binom{\#\text{type}}{k}\cdot 2^{\#\text{rest}}$ — choose the constrained part, then multiply by the full power set of the free part.