Question 1200910
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Question 1
A club with 20 women and 17 men needs to form a committee of size six.
How many committees are possible?
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The seats on the committee aren't named and have equal rank. 
This means the order doesn't matter.
We use the nCr combination formula.


n = 20 women + 17 men = 37 people total
r = 6 selections
n C r = (n!)/(r!(n-r)!)
37 C 6 = (37!)/(6!*(37-6)!)
37 C 6 = (37!)/(6!*31!)
37 C 6 = (37*36*35*34*33*32*31!)/(6!*31!)
37 C 6 = (37*36*35*34*33*32)/(6!)
37 C 6 = (37*36*35*34*33*32)/(6*5*4*3*2*1)
37 C 6 = (1673844480)/(720)
37 C 6 = 2324784



Answer: <font color=red>2,324,784</font>
This is roughly 2.3 million


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Question 2
A club with 20 women and 17 men needs to form a committee of size six.
How many committees are possible if the committee must have three women and three men?
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There are n = 20 women and there are r = 3 selections.
Use the nCr combination formula.
n C r = (n!)/(r!(n-r)!)
20 C 3 = (20!)/(3!*(20-3)!)
20 C 3 = (20!)/(3!*17!)
20 C 3 = (20*19*18*17!)/(3!*17!)
20 C 3 = (20*19*18)/(3!)
20 C 3 = (20*19*18)/(3*2*1)
20 C 3 = (6840)/(6)
20 C 3 = 1140
Take note of the 20*19*18 in the numerator and 3! = 3*2*1 = 6 in the denominator.
There are 1140 ways to pick the three women where order doesn't matter.


Follow similar steps for the men.
n = 17
r = 3
n C r = (n!)/(r!(n-r)!)
17 C 3 = (17!)/(3!*(17-3)!)
17 C 3 = (17!)/(3!*14!)
17 C 3 = (17*16*15*14!)/(3!*14!)
17 C 3 = (17*16*15)/(3!)
17 C 3 = (17*16*15)/(3*2*1)
17 C 3 = (4080)/(6)
17 C 3 = 680
There are 680 ways to pick the three men where order doesn't matter.


Summary:
1140 ways to pick the three women
680 ways to pick the three men


Therefore, 1140*680 = 775,200 is the number of ways to have a committee of 3 women and 3 men. Order doesn't matter.


Answer: <font color=red>775,200</font>


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Question 3
A club with 20 women and 17 men needs to form a committee of size six.
How many committees are possible if the committee must have at least two men?
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Let's consider the event "at most 1 man".
It can be broken down into these two cases:
Case A: 6 women and 0 men on the committee
Case B: 5 women and 1 man on the committee


Case A:
20C6 = 38,760 ways to pick the women
17C0 = 1 way to pick the men (i.e. 1 way to leave the men out entirely)
38,760*1 = 38,760 different ways to do case A.


Case B:
20C5 = 15,504 ways to pick the women
17C1 = 17 ways to pick the men
15,504*17 = 263,568 different ways to do case B.


We have
A+B = 38,760+263,568 = 302,328
ways to have at most 1 man.


This is out of 37C6 = 2,324,784 ways to form the committee (refer to question 1).


2,324,784 - 302,328 = 2,022,456
This is the number of ways to have at least two men. 


The events "at most 1 man" and "at least two men" are complementary. 
They make up all possible outcomes.


Answer: <font color=red>2,022,456</font>


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Question 4
A club with 20 women and 17 men needs to form a committee of size six.
How many committees are possible if the committee must consist of all women or all men?
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Let's determine how many committees are possible if the committee consists of women only.
n = 20 women
r = 6 selections
n C r = (n!)/(r!(n-r)!)
20 C 6 = (20!)/(6!*(20-6)!)
20 C 6 = (20!)/(6!*14!)
20 C 6 = (20*19*18*17*16*15*14!)/(6!*14!)
20 C 6 = (20*19*18*17*16*15)/(6!)
20 C 6 = (20*19*18*17*16*15)/(6*5*4*3*2*1)
20 C 6 = (27907200)/(720)
20 C 6 = 38760


Now let's see how many committees are possible if the committee consists of men only.
n = 17 men
r = 6 selections
n C r = (n!)/(r!(n-r)!)
17 C 6 = (17!)/(6!*(17-6)!)
17 C 6 = (17!)/(6!*11!)
17 C 6 = (17*16*15*14*13*12*11!)/(6!*11!)
17 C 6 = (17*16*15*14*13*12)/(6!)
17 C 6 = (17*16*15*14*13*12)/(6*5*4*3*2*1)
17 C 6 = (8910720)/(720)
17 C 6 = 12376



Answers:
<font color=red>38,760</font> ways to have a committee of all women.
<font color=red>12,376</font> ways to have a committee of all men.
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