Question 1200910: A club with 20 women and 17 men needs to form a committee of size six.
How many committees are possible?
How many committees are possible if the committee must have three women and three men?
How many committees are possible if the committee must have at least two men?
How many committees are possible if the committee must consist of all women or all men?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Question 1
A club with 20 women and 17 men needs to form a committee of size six.
How many committees are possible?
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: 2,324,784
This is roughly 2.3 million
======================================================
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?
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: 775,200
======================================================
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?
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: 2,022,456
======================================================
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?
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:
38,760 ways to have a committee of all women.
12,376 ways to have a committee of all men.
|
|
|
