bucket and dash method for constraint combination problem

jzw wrote:From a group of 7 men and 5 women, how many different committees consisting of 4 men and 2 women can be formed if Jenny and John refuse to be on the committee together?

So first we calculate both not on, then Jenny on but John not, and then John on but Jenny off and add the totals together.

So both not on = 90. That I totally understand. 

But how do I set up the bucket and the dashes for the other 2 possibilities? What am I dividing by and why?

For John not on, I would have thought it would be 6x5x4x3/4x3x2x1 x 5x4/2x1
For Jenny not on, I would have thought it would be 7x6x5x4/4x3x2x1 x 4x3/2x1

Let's forget the added constraint for a moment and calculate the answer.

Number of distinct groups of 4 men = 7C4 = 35
Number of distinct groups of 2 women = 5C2 = 10

Therefore, the total number of distinct groups of men and women = 350

Now, how many of these groups have both Jenny and John?

#Groups of men with John plus 3 other men = 6C3 = 20
#Groups of women with Jenny plus one other woman = 4

Combined, there would have been 80 groups with both.

This leaves 270.

"#Groups of men with John plus 3 other men = 6C3 = 20
#Groups of women with Jenny plus one other woman = 4

Combined, there would have been 80 groups with both."

- can u explain "why" this is so? I need to understand why this works and how to apply this to future problems. Thanks!

jzw wrote:"#Groups of men with John plus 3 other men = 6C3 = 20
#Groups of women with Jenny plus one other woman = 4

Combined, there would have been 80 groups with both."

- can u explain "why" this is so? I need to understand why this works and how to apply this to future problems. Thanks!

Sure! Sorry I didn't go into more detail.

The question states that there are 7 men, and the selected committee will have 4 of these men chosen from the group of 7.

To find the number of groups that John will be a part of, however, we need to take away one of the free slots in the group of four (for John), which leaves us with three spots to filled up by the other six men (again only six because we know that John is already chosen).

So, the number of groups with John is the same as finding the number of distinct groups of three men from a supply of six. This is just 6C3 = 20 groups.

Now we do the same thing to find all of the groups of two women with Jenny in them. This time, there is only one empty slot available, which can be filled by four possible women. In other words, there are 4C1 possible groups of one woman to be paired with Jenny. 4 groups.

Next, we need to determine how many committees of men and women can be created from these 20 groups of men and 4 groups of women. This is actually simple to see (I hope), since each group of men can be paired off with 4 groups of women. So, we simply multiply the 20 * 4 to get 80 distinct committees including both John and Jenny.

Hopefully this makes sense now.

It does!! Thank you!!!!! Now I totally get it. Really appreciate it!!

jzw wrote:It does!! Thank you!!!!! Now I totally get it. Really appreciate it!!

My pleasure.