A class has 10 boys and 12 girls. In how many ways can a committee of four be selected if the committee can have at most two girls?

With no girls

C(10,4) = 210 ways
With one girl
12 C(10,3) = 12*120 = 1440 ways
With two girls
first combinations of 12 girls taken 2 at a time = 12!/[10!*2!) = 12*11/2 = 66
(my table of binomial coefficients only goes to n = 10)
time combinations of ten boys taken 2 at a time = C(10,2) = 45
so 66*45 = 2970
so
2970+1440+210

Well, selecting a committee sounds like a serious task, but let me bring in some humor!

To solve this problem, let's consider different scenarios:

Scenario 1: You select no girls. It's like going to a pizza party and choosing only toppings without the pizza itself! So, you can simply choose 4 boys from the 10 available.

Scenario 2: You select exactly 1 girl. Ah, finding the perfect balance! You can choose 1 girl from the 12 available and 3 boys from the 10 to complete the committee.

Scenario 3: You select exactly 2 girls. It's like having two cute puppies in the committee! In this case, you can choose 2 girls from the 12 available and 2 boys from the 10.

Now, let's calculate the number of possibilities for each scenario:

Scenario 1: Selecting 4 boys from 10 can be calculated using the combination formula: C(10, 4).

Scenario 2: Selecting 1 girl from 12 can be calculated using the combination formula: C(12, 1). Then, selecting 3 boys from 10 can be calculated using the combination formula again: C(10, 3).

Scenario 3: Selecting 2 girls from 12 can be calculated using the combination formula: C(12, 2). Then, selecting 2 boys from 10 can be calculated using the combination formula again: C(10, 2).

Finally, you can find the total number of possibilities by adding up the results from each scenario:

Total Possibilities = C(10, 4) + [C(12, 1) * C(10, 3)] + [C(12, 2) * C(10, 2)]

Now, it's your turn to do the math and find the total number of ways the committee can be selected! Good luck, and remember, even math problems can be funny if you look at them from the right angle!

To find the number of ways a committee of four can be selected, considering at most two girls, we need to break down the problem into three cases:

Case 1: All boys
In this case, we need to select 4 boys from the group of 10. The number of ways to do this is given by the combination formula:
C(10, 4) = 10! / (4!(10-4)!) = 210

Case 2: One girl, three boys
In this case, we need to select 1 girl from the group of 12 and 3 boys from the group of 10. The number of ways to do this is given by the product of the combinations:
C(12, 1) * C(10, 3) = (12! / (1!(12-1)!)) * (10! / (3!(10-3)!)) = 12 * 120 = 1,440

Case 3: Two girls, two boys
In this case, we need to select 2 girls from the group of 12 and 2 boys from the group of 10. The number of ways to do this is given by the product of the combinations:
C(12, 2) * C(10, 2) = (12! / (2!(12-2)!)) * (10! / (2!(10-2)!)) = 66 * 45 = 2,970

Finally, we add up the results from all three cases to get the total number of ways a committee of four can be selected:
210 + 1,440 + 2,970 = 4,620

Therefore, there are 4,620 ways to select a committee of four, considering at most two girls.

no

With no girls

C(10,4) = 210 ways
With one girl
12 C(10,3) = 12*120 = 1440 ways
With two girls
first combinations of 12 girls taken 2 at a time = 12!/[10!*2!) = 12*11/2 = 66
(my table of binomial coefficients only goes to n = 10)
time combinations of ten boys taken 2 at a time = C(10,2) = 45
so 66*45 = 2970
so
2970+1440+210

you just add them up and there you answer