Question

19 multiple choice 2 points
there are 14 girls on the golf team at gu. how many different four - girl teams may be chosen by the coach if jane must be on every team?
1,716
1,001
286
715
20 multiple choice 2 points
you are the coach of a nine - person baseball team. how many different batting orders are possible?
40,320
362,900
5,040
362,880

Explanation:

Question 19

Step1: Determine the number of girls to choose

Since Jane must be on every team, we need to choose \( 4 - 1 = 3 \) more girls from the remaining \( 14 - 1 = 13 \) girls.

Step2: Use the combination formula

The combination formula is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 13 \) and \( k = 3 \).
\[

$$\begin{align*} C(13, 3)&=\frac{13!}{3!(13 - 3)!}\\ &=\frac{13!}{3!×10!}\\ &=\frac{13×12×11×10!}{3×2×1×10!}\\ &=\frac{13×12×11}{6}\\ &=286 \end{align*}$$

\]

Step1: Recognize the permutation problem

We need to find the number of permutations of 9 people (since batting order is a permutation, as order matters). The permutation formula is \( P(n, n)=\frac{n!}{(n - n)!}=n! \).

Step2: Calculate \( 9! \)

\[

$$\begin{align*} 9!&=9×8×7×6×5×4×3×2×1\\ &=362880 \end{align*}$$

\]

Answer:

286

Question 20