SOLUTION: A committee of two men and three women is to be formed from 6 men and 4 women. How many different committees can be formed if;
(i) there are no restrictions;
(ii) a particular
Algebra.Com
Question 1124688: A committee of two men and three women is to be formed from 6 men and 4 women. How many different committees can be formed if;
(i) there are no restrictions;
(ii) a particular man and a particular woman cannot serve on the same committee.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Part (i)
How many ways are there to pick the two male candidates?
n = 6 men total to pick from
r = 2 men selected from the pool of six
n C r = (n!)/(r!(n-r)!) <<--- this is the combination formula
6 C 2 = (6!)/(2!*(6-2)!)
6 C 2 = (6!)/(2!*4!)
6 C 2 = (6*5*4*3*2*1)/((2*1)*(4*3*2*1))
6 C 2 = (720)/((2)*(24))
6 C 2 = (720)/(48)
6 C 2 = 15
There are 15 ways to pick the two male selections
--------------------------------------------------
How many ways are there to pick the three female candidates?
n = 4 women total to pick from
r = 3 women are to be selected from the pool of four
n C r = (n!)/(r!(n-r)!)
4 C 3 = (4!)/(3!*(4-3)!)
4 C 3 = (4!)/(3!*1!)
4 C 3 = (4*3*2*1)/((3*2*1)*(1))
4 C 3 = (24)/((6)*(1))
4 C 3 = (24)/(6)
4 C 3 = 4
There are 4 ways to pick the three female selections.
Note: this is the same as asking "how many ways are there to not select a female candidate" since one female is left out from the other group of 3.
This explains why 4 C (4-1) = 4 or 4 C 3 = 4 which in general means n C (n-1) = n.
--------------------------------------------------
To summarize so far: We have 15 ways to pick the two men, and 4 ways to pick the three women.
This means that 15*4 = 60 represents the total number of ways to pick the entire committee consisting of two men and three women.
There are no restrictions on this group other than the fact that there are 2 men and 3 women.
Answer: 60
=======================================================================================
=======================================================================================
Part (ii)
Label the set of males as: {M1, M2, M3, M4, M5, M6}
Label the set of females as: {F1, F2, F3, F4}
Let's say that male candidate M1 and female candidate F1 cannot serve together. The choices M1 and F1 are completely arbitrary.
If person M1 is chosen, then person F1 cannot be selected as well. We need to figure out how many times M1 and F1 show up in the same group, so we know what groups to avoid.
These groups will be subtracted off later. To find the number of groups where M1 and F1 are together, let's reserve 2 slots for M1 and F1 out of 5 slots total (Slot A,B,C,D,E)
Slot A is a reserved slot for person M1
Slot B is a reserved slot for person F1
Slot C is a slot for a male candidate
Slot D is a slot for a female candidate
Slot E is a slot for a female candidate
--------------------------------------------------
Focus on slots C through E:
Slot C has 5 choices to pick from because there are 5 males left over (after M1 is picked)
Slot D has 3 choices to pick from. The pool {F1, F2, F3, F4} becomes {F2, F3, F4} after F1 is chosen
Slot E has 2 choices to pick from. Whoever gets in slot D is taken out of the pool of candidates to drop from 3 to 3-1 = 2.
We have 3*2 = 6 ways to pick the other two females, but since order doesn't matter this means we really have 6/2 = 3 different ways to pick the other females. Then consider the other 5 male choices (slot C) and we have 5*3 = 15 ways to form the committee in the form {M1, F1, X, Y, Z} where X,Y,Z are placeholders for the other candidates chosen.
Here are the 15 ways to pick the committee with M1 and F1 on together
{M1,F1,M2,F2,F3} 1
{M1,F1,M2,F2,F4} 2
{M1,F1,M2,F3,F4} 3
--------
{M1,F1,M3,F2,F3} 4
{M1,F1,M3,F2,F4} 5
{M1,F1,M3,F3,F4} 6
--------
{M1,F1,M4,F2,F3} 7
{M1,F1,M4,F2,F4} 8
{M1,F1,M4,F3,F4} 9
--------
{M1,F1,M5,F2,F3} 10
{M1,F1,M5,F2,F4} 11
{M1,F1,M5,F3,F4} 12
--------
{M1,F1,M6,F2,F3} 13
{M1,F1,M6,F2,F4} 14
{M1,F1,M6,F3,F4} 15
We have five groups with 3 blocks per group, so there are 15 total shown above
The numbers in blue are simply used to count out the possibilities shown above
--------------------------------------------------
We have 15 ways to pick a group that will be guaranteed to have M1 and F1 serving together on the same committee
We want to avoid this scenario, so we subtract it from the total 60 (found back in part (i)) to get 60-15 = 45.
There are 45 ways to pick a group that does not have M1 and F1 both serving together.
Either M1 is on the committee, F1 is on the committee, or neither person is.
Answer: 45
RELATED QUESTIONS
From a group of 6 women and 8 men, a committee consisting of 4 men and 3 women is to be... (answered by ikleyn)
How many different committees can be formed from 9 women and 8 men if the committee... (answered by ikleyn)
There are 12 women and 5 men. A committee of 7 women and 4 men is to be chosen. How many... (answered by SUNIL1987MAURYA)
Six men and eight women volunteered to serve on the committee. How many different... (answered by jorel1380)
From a group of 5 women and 7 men, how many different committees consisting of 2 women... (answered by jrfrunner)
from a group of 5 men and 7 women, a committee of 2men and 3 women is to be formed. how (answered by sudhanshu_kmr)
A committee is to have four members. There are seven men and six women available to serve (answered by stanbon)
From a group of 6 men and 4 women, how many committees of 2 men and 3 women can be... (answered by solver91311)
from a group of 5 men and 6 women how many committees of 4 men and 3 women can be... (answered by ewatrrr)