The camera club has 5 members, and the math club has 8 members. There is only one member common to both clubs. In how many ways could a committee of 4 people be formed with at least 1 member from each group?

Let's say Bob is in both clubs. There are 11 other people.

Two cases: Bob is on the committee, or not:

a) Bob included:
First choose him, then choose any 3 of the other 11, since both clubs are represented by Bob:
11C3 ways to do that.

b) Bob excluded:
Now start with 11C4 (committees made up of the other 11 people).
Subtract 4C4 = 1 way with all 4 others from the camera club,
and 7C4 = 35 ways with 4 others from the math club.

11C3 + 11C4 - 1 - 7C4 = 165 + 330 - 1 - 35 = 459

Or we could say:
there are 12C4 total committees ( = 495 = 165 + 300 )
not worrying about Bob, and then just subtract the 1 which is the 4 camera club members
who are not Bob and the 35, which is the same for the math club.

Well, first of all, let's address the elephant in the room: Why did the camera club and math club have only one member in common? Did they have a secret feud going on? Maybe the camera club shuttered every time a math member approached, who knows?

Now, back to the question at hand! Since we need at least one member from each club, we can choose 1 member from the camera club in 5 ways (since it has 5 members) and choose 1 member from the math club in 8 ways (since it has 8 members). That gives us a total of 5 x 8 = 40 ways to select the initial 2 members.

Now we need to choose 2 more members from the remaining members in both clubs. From the camera club, we have 5 - 1 (since we already selected 1 member) = 4 remaining members to choose from. And from the math club, we have 8 - 1 = 7 remaining members to choose from.

To calculate the number of ways to select the remaining 2 members, we can use combinations. So we have C(4, 2) for the camera club and C(7, 2) for the math club, which equals 6 and 21, respectively.

Finally, multiplying all the possibilities together gives us 40 x 6 x 21 = 5040 ways to form a committee of 4 people with at least 1 member from each group.

Remember, though, committees are serious business. Handle with care, use humor responsibly!

To form a committee of 4 people with at least 1 member from each group (camera club and math club), we can consider the following cases:

1. Selecting 1 member from the camera club and 3 members from the math club:
- The number of ways to select 1 member from the camera club: 5C1 = 5.
- The number of ways to select 3 members from the math club: 8C3 = 56.
- Total number of ways for this case: 5 * 56 = 280.

2. Selecting 2 members from the camera club and 2 members from the math club:
- The number of ways to select 2 members from the camera club: 5C2 = 10.
- The number of ways to select 2 members from the math club: 8C2 = 28.
- Total number of ways for this case: 10 * 28 = 280.

3. Selecting 3 members from the camera club and 1 member from the math club:
- The number of ways to select 3 members from the camera club: 5C3 = 10.
- The number of ways to select 1 member from the math club: 8C1 = 8.
- Total number of ways for this case: 10 * 8 = 80.

4. Selecting 4 members from the camera club and no member from the math club:
- The number of ways to select 4 members from the camera club: 5C4 = 5.

Therefore, the total number of ways to form a committee of 4 people with at least 1 member from each group is: 280 + 280 + 80 + 5 = 645 ways.

To find the number of ways to form a committee of 4 people with at least 1 member from each group, we can use the principle of inclusion-exclusion.

First, we need to calculate the total number of ways to form a committee without any restrictions. For this, we need to choose 4 members from a total of 5+8=13 members. This can be done using combinations, denoted by "C". Therefore, the total number of ways to form the committee is C(13, 4) = 715.

Next, we need to subtract the number of committees that only have members from the camera club or only have members from the math club, as these cases do not satisfy the requirement of having at least 1 member from each group.

In the first case, when the committee is formed with only camera club members, we need to choose 4 members from a total of 5 camera club members. This can be calculated as C(5, 4) = 5.

In the second case, when the committee is formed with only math club members, we need to choose 4 members from a total of 8 math club members. This can be calculated as C(8, 4) = 70.

However, we have subtracted twice the committees with only one club's members. We need to add the number of committees that overlap. Since there is only 1 member common to both clubs, we need to choose 3 members from a total of 4 remaining members from the camera club (5-1=4). This can be calculated as C(4, 3) = 4.

Therefore, the number of committees formed that have at least 1 member from each group can be calculated as:
Total number of ways - (Number of committees with only camera members + Number of committees with only math members - Number of committees with members from both clubs)
= 715 - (5 + 70 - 4)
= 644.

Thus, there are 644 ways to form a committee of 4 people with at least 1 member from each group.