The camera club has 5 members and the mathematics club has 8. There is only one member common to both clubs. In how many ways could a committee be formed with at least one member from each club?
How large is the committee ?
oops. It's a committee of 4 people
use the binomial theorem to expand(3x-2y)4
To find the number of ways a committee can be formed with at least one member from each club, we can use the concept of combinations.
First, let's consider forming a committee without any restrictions. We can choose members from the camera club in 2^5 = 32 ways (since each member has two choices: either to be included or not). Similarly, we can choose members from the mathematics club in 2^8 = 256 ways.
However, this includes cases where we have committees with members only from one club or neither club. So, we need to subtract those cases.
To form a committee with only members from the camera club, there is only one possible way (all 5 members). Similarly, forming a committee with only members from the mathematics club has only one possible way (all 8 members).
To form a committee with neither club's members, there is only one possibility (no member at all).
Therefore, the number of ways to form a committee without any restrictions is 32 * 256 = 8,192.
Next, we need to consider the cases where there is at least one member from both clubs. We know that there is only one member common to both clubs, so we can choose this member in 1 way.
Now, we need to choose the remaining members from the camera club and the remaining members from the mathematics club. These remaining members can include none, some, or all of the members from each club.
For the camera club, we have 5 - 1 = 4 remaining members to choose from, and for the mathematics club, we have 8 - 1 = 7 remaining members to choose from.
We can choose the remaining camera club members in 2^4 = 16 ways (since each remaining member has two choices: either to be included or not). Similarly, we can choose the remaining mathematics club members in 2^7 = 128 ways.
To calculate the total number of valid committees, we can multiply the possibilities together: 1 * 16 * 128 = 2,048.
Therefore, there are 2,048 ways to form a committee with at least one member from each club.
So the only case we DON'T want are committees consisting of all camera members and all math club members.
with out any restrictions we choose 4 from the 13
= C(13,4) = 715
all camera people = C(5,4) = 5
all math club people = C(8,4) = 70
so number with at least one from each club
= 715 - 5 - 70 = 640