Respuesta :
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 [tex]_6C_2 = 15[/tex] different ways to pick the two boys for the committee.
Here are the steps as to how I got that value of 15
[tex]_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\\\\[/tex]
Through similar steps, you should find that [tex]_8C_3 = 56[/tex]. 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.
Answer: 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 [tex]_6C_3 = 20[/tex] ways to pick the boys and [tex]_8C_2 = 28[/tex] 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 [tex]_6C_4 = 15[/tex] ways to pick the boys and [tex]_8C_1 = 8[/tex] 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 [tex]_6C_5 = 6[/tex]
We add up the results of each bullet point
840+560+120+6 = 1526
Answer: 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 [tex]_{12}C_3 = 220[/tex]
Answer: 220
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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 [tex]_8C_3 = 56[/tex] ways to pick the girls and [tex]_6C_2 = 15[/tex] 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 [tex]_8C_4 = 70[/tex] ways to pick the girls and [tex]_6C_1 = 6[/tex] ways to pick the boy. That leads to 70*6 = 420 different combos for this scenario.
- If there are 5 girls, then there are [tex]_8C_5 = 56[/tex] ways to pick the girls and [tex]_6C_0 = 1[/tex] 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
