Mathematics, 25.01.2022 02:50 dominikkovar

8. David and Susan belong to a math club at their school. There are 6 boys and 8 girls in the club. How many different ways can a 5-person committee be selected from the 14
club members under each of the following conditions?
a. There must be 2 boys and 3 girls
b. There must be at least 2 boys.
c. David (a boy) and Susan (a girl) must be on the committee
d. There must be more girls than boys

Answers: 3

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Answer from: selena5713

Part A

We'll use the nCr combination formula. This is because order doesn't matter on a committee. Each person has equal rank.

We have n = 6 boys and r = 2 slots for them. This means there are _6C_2 = 15 different ways to pick the two boys for the committee.

Here are the steps as to how I got that value of 15

$_n C_r = \frac{n!}{r!*(n-r)!}\\\\
_6 C_2 = \frac{6!}{2!*(6-2)!}\\\\
_6 C_2 = \frac{6!}{2!*4!}\\\\
_6 C_2 = \frac{6*5*4*3*2*1}{(2*1)*(4*3*2*1)}\\\\
_6 C_2 = \frac{720}{48}\\\\
_6 C_2 = 15\\\\$

Through similar steps, you should find that _8C_3 = 56 . There are 56 ways to pick the three girls from 8 girls in the math club. This time I used n = 8 and r = 3 in the formula.

We found there are 15 ways to pick the boys and 56 ways to pick the girls. Overall, there are 15*56 = 840 ways to pick the five students such that there are 2 boys and 3 girls.

840

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Part B

The phrasing "at least 2" means "2 or more".

The number of boys that could be on the committee is a number from the set {2,3,4,5}.

If the committee has 2 boys and 3 girls, then there are 840 ways to form the committee. This value was calculated earlier in part A.If there are 3 boys and 2 girls, then there are _6C_3 = 20

ways to pick the boys and _8C_2 = 28

ways to pick the girls. That gives 20*28 = 560 ways to pick this combination of students.If there are 4 boys and 1 girl, then there are _6C_4 = 15

ways to pick the boys and _8C_1 = 8

ways to pick the girls. This yields 15*8 = 120 different combos.Lastly, if there are 5 boys and 0 girls, then there are 6 ways to form this group (since there are 6 ways to leave a certain boy out of the group). Or you could compute _6C_5 = 6

We add up the results of each bullet point

840+560+120+6 = 1526

1526

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Part C

Davin and Susan must be on the committee. So the 5 initial seats drop to 5-2 = 3 seats left after locking in those two people.

Afterward, there are no restrictions on the 3 remaining seats.

There are n = 14-2 = 12 students left and r = 3 slots to fill.

Use the nCr combination formula to get $_{12}C_3 = 220$

220

==========================================================

Part D

If there must be more girls than boys, then the number of girls could be from the set {3,4,5}

If there are 3 girls and 2 boys, then there are _8C_3 = 56

ways to pick the girls and _6C_2 = 15

ways to pick the boys. That gives 56*15 = 840 combos (see part A).If there are 4 girls and 1 boy, then there are _8C_4 = 70

ways to pick the girls and _6C_1 = 6

ways to pick the boy. That leads to 70*6 = 420 different combos for this scenario.If there are 5 girls, then there are _8C_5 = 56

ways to pick the girls and _6C_0 = 1

way to have a group of no boys. Therefore we have 56*1 = 56 ways to have a committee of all girls.

Add up the results

840+420+56 = 1316

1316

Answer from: Quest

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

Answer from: Quest

answer: about 22.67

step-by-step explanation:

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8. David and Susan belong to a math club at their school. There are 6 boys and 8 girls in the club...

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