A committee must be formed with $3$ teachers and $7$ students. If there are $9$ teachers to choose from, and $15$ students, how many different ways could the committee be made? $\newline$ Answer:

View step-by-step help

Home

Math Problems

Geometry

Counting principle

Full solution

Q. A committee must be formed with

3

teachers and

7

students. If there are

9

teachers to choose from, and

15

students, how many different ways could the committee be made?

\newline

Answer:

Calculate Teachers Combination: To determine the number of different ways to form the committee, we need to calculate the combinations of teachers and students separately and then multiply them together. For the teachers, we will use the combination formula which is $C(n, k) = \frac{n!}{k!(n - k)!}$ , where $n$ is the total number of items to choose from, $k$ is the number of items to choose, and $＂!＂$ denotes factorial.
Calculate Students Combination: First, we calculate the number of ways to choose $3$ teachers out of $9$ . Using the combination formula, we get $C(9, 3) = \frac{9!}{3!(9 - 3)!}$ .
Calculate Total Ways: Calculating the factorials, we have $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ . We can simplify $9! / 6!$ by canceling out the common terms, which leaves us with $9 \times 8 \times 7$ .
Calculate Total Ways: Calculating the factorials, we have $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ . We can simplify $9! / 6!$ by canceling out the common terms, which leaves us with $9 \times 8 \times 7$ .Now, we calculate $9 \times 8 \times 7 / 3! = 9 \times 8 \times 7 / (3 \times 2 \times 1) = 9 \times 4 \times 7 = 252$ . So, there are $252$ ways to choose $3$ teachers from $9$ .
Calculate Total Ways: Calculating the factorials, we have $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ . We can simplify $9! / 6!$ by canceling out the common terms, which leaves us with $9 \times 8 \times 7$ .Now, we calculate $9 \times 8 \times 7 / 3! = 9 \times 8 \times 7 / (3 \times 2 \times 1) = 9 \times 4 \times 7 = 252$ . So, there are $252$ ways to choose $3$ teachers from $9$ .Next, we calculate the number of ways to choose $7$ students out of $3! = 3 \times 2 \times 1$ $0$ . Using the combination formula, we get $3! = 3 \times 2 \times 1$ $1$ .
Calculate Total Ways: Calculating the factorials, we have $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ . We can simplify $9! / 6!$ by canceling out the common terms, which leaves us with $9 \times 8 \times 7$ .Now, we calculate $9 \times 8 \times 7 / 3! = 9 \times 8 \times 7 / (3 \times 2 \times 1) = 9 \times 4 \times 7 = 252$ . So, there are $252$ ways to choose $3$ teachers from $9$ .Next, we calculate the number of ways to choose $7$ students out of $3! = 3 \times 2 \times 1$ $0$ . Using the combination formula, we get $3! = 3 \times 2 \times 1$ $1$ .Calculating the factorials, we have $3! = 3 \times 2 \times 1$ $2$ , $3! = 3 \times 2 \times 1$ $3$ , and $3! = 3 \times 2 \times 1$ $4$ . We can simplify $3! = 3 \times 2 \times 1$ $5$ by canceling out the common terms, which leaves us with $3! = 3 \times 2 \times 1$ $6$ .
Calculate Total Ways: Calculating the factorials, we have $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ . We can simplify $9! / 6!$ by canceling out the common terms, which leaves us with $9 \times 8 \times 7$ .Now, we calculate $9 \times 8 \times 7 / 3! = 9 \times 8 \times 7 / (3 \times 2 \times 1) = 9 \times 4 \times 7 = 252$ . So, there are $252$ ways to choose $3$ teachers from $9$ .Next, we calculate the number of ways to choose $7$ students out of $3! = 3 \times 2 \times 1$ $0$ . Using the combination formula, we get $3! = 3 \times 2 \times 1$ $1$ .Calculating the factorials, we have $3! = 3 \times 2 \times 1$ $2$ , $3! = 3 \times 2 \times 1$ $3$ , and $3! = 3 \times 2 \times 1$ $4$ . We can simplify $3! = 3 \times 2 \times 1$ $5$ by canceling out the common terms, which leaves us with $3! = 3 \times 2 \times 1$ $6$ .Now, we calculate $3! = 3 \times 2 \times 1$ $7$ . We can cancel out the $7$ from the numerator and denominator, and then simplify further.
Calculate Total Ways: Calculating the factorials, we have $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ . We can simplify $9! / 6!$ by canceling out the common terms, which leaves us with $9 \times 8 \times 7$ .Now, we calculate $9 \times 8 \times 7 / 3! = 9 \times 8 \times 7 / (3 \times 2 \times 1) = 9 \times 4 \times 7 = 252$ . So, there are $252$ ways to choose $3$ teachers from $9$ .Next, we calculate the number of ways to choose $7$ students out of $3! = 3 \times 2 \times 1$ $0$ . Using the combination formula, we get $3! = 3 \times 2 \times 1$ $1$ .Calculating the factorials, we have $3! = 3 \times 2 \times 1$ $2$ , $3! = 3 \times 2 \times 1$ $3$ , and $3! = 3 \times 2 \times 1$ $4$ . We can simplify $3! = 3 \times 2 \times 1$ $5$ by canceling out the common terms, which leaves us with $3! = 3 \times 2 \times 1$ $6$ .Now, we calculate $3! = 3 \times 2 \times 1$ $7$ . We can cancel out the $7$ from the numerator and denominator, and then simplify further.After simplification, we get $3! = 3 \times 2 \times 1$ $9$ . We can cancel out the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $0$ in the denominator with the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $1$ in the numerator to get $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $2$ , and the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $3$ with the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $4$ to get $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $2$ .
Calculate Total Ways: Calculating the factorials, we have $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ . We can simplify $9! / 6!$ by canceling out the common terms, which leaves us with $9 \times 8 \times 7$ .Now, we calculate $9 \times 8 \times 7 / 3! = 9 \times 8 \times 7 / (3 \times 2 \times 1) = 9 \times 4 \times 7 = 252$ . So, there are $252$ ways to choose $3$ teachers from $9$ .Next, we calculate the number of ways to choose $7$ students out of $3! = 3 \times 2 \times 1$ $0$ . Using the combination formula, we get $3! = 3 \times 2 \times 1$ $1$ .Calculating the factorials, we have $3! = 3 \times 2 \times 1$ $2$ , $3! = 3 \times 2 \times 1$ $3$ , and $3! = 3 \times 2 \times 1$ $4$ . We can simplify $3! = 3 \times 2 \times 1$ $5$ by canceling out the common terms, which leaves us with $3! = 3 \times 2 \times 1$ $6$ .Now, we calculate $3! = 3 \times 2 \times 1$ $7$ . We can cancel out the $7$ from the numerator and denominator, and then simplify further.After simplification, we get $3! = 3 \times 2 \times 1$ $9$ . We can cancel out the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $0$ in the denominator with the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $1$ in the numerator to get $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $2$ , and the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $3$ with the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $4$ to get $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $2$ .The final calculation is $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $6$ . So, there are $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $7$ ways to choose $7$ students from $3! = 3 \times 2 \times 1$ $0$ .
Calculate Total Ways: Calculating the factorials, we have $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ . We can simplify $9! / 6!$ by canceling out the common terms, which leaves us with $9 \times 8 \times 7$ .Now, we calculate $9 \times 8 \times 7 / 3! = 9 \times 8 \times 7 / (3 \times 2 \times 1) = 9 \times 4 \times 7 = 252$ . So, there are $252$ ways to choose $3$ teachers from $9$ .Next, we calculate the number of ways to choose $7$ students out of $3! = 3 \times 2 \times 1$ $0$ . Using the combination formula, we get $3! = 3 \times 2 \times 1$ $1$ .Calculating the factorials, we have $3! = 3 \times 2 \times 1$ $2$ , $3! = 3 \times 2 \times 1$ $3$ , and $3! = 3 \times 2 \times 1$ $4$ . We can simplify $3! = 3 \times 2 \times 1$ $5$ by canceling out the common terms, which leaves us with $3! = 3 \times 2 \times 1$ $6$ .Now, we calculate $3! = 3 \times 2 \times 1$ $7$ . We can cancel out the $7$ from the numerator and denominator, and then simplify further.After simplification, we get $3! = 3 \times 2 \times 1$ $9$ . We can cancel out the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $0$ in the denominator with the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $1$ in the numerator to get $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $2$ , and the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $3$ with the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $4$ to get $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $2$ .The final calculation is $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $6$ . So, there are $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $7$ ways to choose $7$ students from $3! = 3 \times 2 \times 1$ $0$ .To find the total number of different ways to form the committee, we multiply the number of ways to choose the teachers by the number of ways to choose the students: $252$ (ways to choose teachers) $9! / 6!$ $1$ $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $7$ (ways to choose students).
Calculate Total Ways: Calculating the factorials, we have $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ . We can simplify $9! / 6!$ by canceling out the common terms, which leaves us with $9 \times 8 \times 7$ .Now, we calculate $9 \times 8 \times 7 / 3! = 9 \times 8 \times 7 / (3 \times 2 \times 1) = 9 \times 4 \times 7 = 252$ . So, there are $252$ ways to choose $3$ teachers from $9$ .Next, we calculate the number of ways to choose $7$ students out of $3! = 3 \times 2 \times 1$ $0$ . Using the combination formula, we get $3! = 3 \times 2 \times 1$ $1$ .Calculating the factorials, we have $3! = 3 \times 2 \times 1$ $2$ , $3! = 3 \times 2 \times 1$ $3$ , and $3! = 3 \times 2 \times 1$ $4$ . We can simplify $3! = 3 \times 2 \times 1$ $5$ by canceling out the common terms, which leaves us with $3! = 3 \times 2 \times 1$ $6$ .Now, we calculate $3! = 3 \times 2 \times 1$ $7$ . We can cancel out the $7$ from the numerator and denominator, and then simplify further.After simplification, we get $3! = 3 \times 2 \times 1$ $9$ . We can cancel out the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $0$ in the denominator with the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $1$ in the numerator to get $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $2$ , and the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $3$ with the $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $4$ to get $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $2$ .The final calculation is $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $6$ . So, there are $(9 - 3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $7$ ways to choose $7$ students from $3! = 3 \times 2 \times 1$ $0$ .To find the total number of different ways to form the committee, we multiply the number of ways to choose the teachers by the number of ways to choose the students: $252$ (ways to choose teachers) $9! / 6!$ $1$ (ways to choose students).Multiplying these together, we get $9! / 6!$ $2$ . Therefore, there are $9! / 6!$ $3$ different ways to form the committee.