Problem Solving

A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?
i found the answer ( my answer) is 90...this is how i did it
3 couples of married people : AabBcC, there are 3 ways to choose 1 in 3 couples
for 2 left,
+ choose 1b,1c= 3*2*2*2=24
+choose 1b or 1 c and choose 1 from singles men=3*2*2*4=48
+choose 2 from single men= 6*3=18

but the answer i got is not 90 so could you guys please help? and tell me the fastest way to solve this problem?

ohh got it, i got it...
i could do like this: choose 4 from 10 people first so there could be 210
then choose 2 couples from 3 couples: = 3
the answer is 10C4-2C3=207
please delete this post...

3 couples and 4 men. atmost 1 couple = 0 couple or 1 couple. = total - 2 couple...
total 10C4 - 3C2= 207...

cans wrote:3 couples and 4 men. atmost 1 couple = 0 couple or 1 couple. = total - 2 couple...
total 10C4 - 3C2= 207...

I think this is the best and fastest way to solve this problem.. thanks Cans

The following is the wrong way to do it, but it confirms that 207 can be reached by adding up all the cases:

Call the people AABBCCSSSS

1. Couple + 2 others
3 choices for couple
8C2 = 28 ways to choose the other 2, but we must subtract 2 because we can't choose the two couples, so 26 ways to choose the other 2. 3 * 26 = 78.

2. ABCS
2 choices for A, B, C
4 choices for S
2 * 2 * 2 * 4 = 32

3. 1 person from each of two couples and 2 single people (e.g. ABSS)
6C2 = 15 ways to choose 2 people from AABBCC but we must subtract 3 because we can't choose the couples, so 12 ways to choose 2 people from the couples.
4C2 = 6 ways to choose SS.
12 * 6 = 72.

4. 1 person from a couple, 3 single people
6 choices from the person from couple
4C3 = 4 ways to choose SSS
6 * 4 = 24

5. 4 single people
1 way to do this

78 + 32 + 72 + 24 + 1 = 207

What's wrong with the following approach:

# of ways to select 4 ppl from:

4 single men= 4C4=1
3 single men and any one from married couple= 4C3*6C1 = 24
2 single men and any two from married couple= 4C2*6C1*4C1= 144
1 single men and any three from married couple= 4C1*6C1*4C1*2C1= 192

Also, accounting for atmost 1 married couple= 3C2*(8C2-2) = 78

Combining:
1+24+144+192+78 which is way too much than required.

Please help.

Thank you.

parveen110 wrote:What's wrong with the following approach:

# of ways to select 4 ppl from:

4 single men= 4C4=1
3 single men and any one from married couple= 4C3*6C1 = 24
2 single men and any two from married couple= 4C2*6C1*4C1= 144
The portion in red is incorrect.
If your intention is to select a COMBINATION of 2 married people who are not married to each other, then the ORDER of the 2 married people doesn't matter.
Thus, we must divide 6C1*4C1 by the the number of ways the 2 married people can be arranged (2!):
4C2 * (6C1*4C1)/2! = 72.

1 single men and any three from married couple= 4C1*6C1*4C1*2C1= 192
The portion in red is incorrect.
If your intention is to select a COMBINATION of 3 married people who are not married to each other, then the ORDER of the 3 married people doesn't matter.
Thus, we must divide 6C1*4C1*2C1 by the the number of ways the 3 married people can be arranged (3!):
4C1 * (6C1*4C1*2C1)/3! = 32.

Also, accounting for at most 1 married couple= 3C2*(8C2-2) = 78

Combining:
1+24+144+192+78 which is way too much than required.
Adding together the revised totals, we get:
1+24+72+32+78 = 207.

Please help.

Thank you.

Please see my notes in red above.
The following approach is similar to yours but perhaps a bit more straightforward:

Case 1: 4 single men
Number of ways to choose 4 single men from 4 options = 4C4 = (4*3*2*1)/(4*3*2*1) = 1.

Case 2: 3 single men and 1 married person
Number of ways to choose 3 single men from 4 options = 4C3 = (4*3*2)/(3*2*1) = 4.
Number of ways to choose 1 married person from 6 options = 6C1 = 6/1 = 6.
To combine these options, we multiply:
4*6 = 24.

Case 3: 2 single men and 2 married people
Number of ways to choose 2 single men from 4 options = 4C2 = (4*3)/(2*1) = 6.
Number of ways to choose 2 married people from 6 options = 6C2 = (6*5)/(2*1) = 15.
To combine these options, we multiply:
6*15 = 90.

Case 4: 1 single men and 3 married people
Number of ways to choose 1 single man from 4 options = 4C1 = 4/1 = 4.
Number of ways to choose 3 married people from 6 options = 6C3 = (6*5*4)/(3*2*1) = 20.
To combine these options, we multiply:
4*20 = 80.

Case 5: 4 married people
Number of ways to choose 4 married people from 6 options = 6C4 = 15.
Of these 15 combinations, we must subtract those consisting of 2 married couples.
Number of ways to choose 2 married couples from 3 options = 3C2 = 3.
Subtracting the 3 disallowed combinations, we get:
15-3 = 12.

Adding together the 5 cases, we get:
1 + 24 + 90 + 80 + 12 = 207.