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The homework for English class was to write a poem. The teacher wants to ask 4 students, 2 boys and 2 sirls, to read their poems for the class. If there are 10 boys and 15 girls, how many different combinations of 2 boys and 2 girls can the teacher select?

The homework for English class was to write a poem. The teacher wants to ask $4$ students, $2$ boys and $2$ sirls, to read their poems for the class. If there are $10$ boys and $15$ girls, how many different combinations of $2$ boys and $2$ girls can the teacher select?

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Sine and cosine of complementary angles

Full solution

Q. The homework for English class was to write a poem. The teacher wants to ask

4

students,

2

boys and

2

sirls, to read their poems for the class. If there are

10

boys and

15

girls, how many different combinations of

2

boys and

2

girls can the teacher select?

Boys Combination Calculation: To find the number of combinations of $2$ boys out of $10$ , we use the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items, $k$ is the number of items to choose, and ! denotes factorial. $\newline$ For $2$ boys out of $10$ , the calculation is $C(10, 2) = \frac{10!}{2!(10-2)!}$ .
Boys Combination Result: Calculating $C(10, 2)$ gives us $\frac{10!}{2! \times 8!} = \frac{(10 \times 9)}{(2 \times 1)} = 45$ . $\newline$ There are $45$ different ways to choose $2$ boys out of $10$ .
Girls Combination Calculation: Next, we find the number of combinations of $2$ girls out of $15$ using the same combination formula. $\newline$ For $2$ girls out of $15$ , the calculation is $C(15, 2) = \frac{15!}{(2!(15-2)!)}$ .
Girls Combination Result: Calculating $C(15, 2)$ gives us rac{15!}{2! imes 13!} = rac{(15 imes 14)}{(2 imes 1)} = 105. $\newline$ There are $105$ different ways to choose $2$ girls out of $15$ .
Total Combinations Calculation: To find the total number of combinations of $2$ boys and $2$ girls, we multiply the number of combinations for boys by the number of combinations for girls. $\newline$ The calculation is $45 \times 105$ .
Total Combinations Result: Calculating $45 \times 105$ gives us $4725$ . There are $4725$ different combinations of $2$ boys and $2$ girls that the teacher can select.

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