A class consists of 12 boys and 8 girls. In how many ways can a committee of 6 be formed if:
there must be at least 1 boy and 1 girl?
I got 896 at the end but not sure if it is right.
no boys ... 8C6 = 28
no girls ... 12C6 = 924
any combination ... 20C6 = 38760
at least one of each ... 38760 - 924 - 28 = ?
Why did the committee get together? Because they wanted to find the perfect balance of boys and girls, of course! Let's see if we can figure out the number of ways to do that.
First, we need to select at least 1 boy and 1 girl for the committee. So we have 12 boys to choose from and 8 girls to choose from. That's 12 choices for the first boy, and then 8 choices for the first girl. Multiplying those choices together, we have 12 * 8 = 96 ways to select the first boy and first girl.
Now, we need to choose the remaining 4 members of the committee. Since we've already chosen 1 boy and 1 girl, we have 11 boys left to choose from and 7 girls left to choose from. That's 11 choices for the second boy, and then 7 choices for the second girl. Multiplying those choices together, we have 11 * 7 = 77 ways to select the second boy and second girl.
For the third member of the committee, we have 10 boys left to choose from and 6 girls left to choose from. That's 10 choices for the third boy, and then 6 choices for the third girl. Multiplying those choices together, we have 10 * 6 = 60 ways to select the third boy and third girl.
Now, for the fourth member, we have 9 boys left to choose from and 5 girls left to choose from. That's 9 choices for the fourth boy, and then 5 choices for the fourth girl. Multiplying those choices together, we have 9 * 5 = 45 ways to select the fourth boy and fourth girl.
Moving on to the fifth member, we have 8 boys left to choose from and 4 girls left to choose from. That's 8 choices for the fifth boy, and then 4 choices for the fifth girl. Multiplying those choices together, we have 8 * 4 = 32 ways to select the fifth boy and fifth girl.
Finally, for the sixth member, we have 7 boys left to choose from and 3 girls left to choose from. That's 7 choices for the sixth boy, and then 3 choices for the sixth girl. Multiplying those choices together, we have 7 * 3 = 21 ways to select the sixth boy and sixth girl.
To get the total number of ways to form the committee, we need to multiply all these individual choices together: 96 * 77 * 60 * 45 * 32 * 21 = 50,138,112.
So, it seems that there are 50,138,112 ways to form a committee of 6 with at least 1 boy and 1 girl. But remember, this is not a laughing matter! It's just math, plain and simple.
To find the number of ways to form a committee of 6 members with at least 1 boy and 1 girl, we need to consider different scenarios:
1. Selecting 1 girl and 5 boys:
- Number of ways to select 1 girl from 8 girls: C(8, 1) = 8
- Number of ways to select 5 boys from 12 boys: C(12, 5) = 792
- Total number of ways for this scenario = 8 * 792 = 6,336
2. Selecting 2 girls and 4 boys:
- Number of ways to select 2 girls from 8 girls: C(8, 2) = 28
- Number of ways to select 4 boys from 12 boys: C(12, 4) = 495
- Total number of ways for this scenario = 28 * 495 = 13,860
3. Selecting 3 girls and 3 boys:
- Number of ways to select 3 girls from 8 girls: C(8, 3) = 56
- Number of ways to select 3 boys from 12 boys: C(12, 3) = 220
- Total number of ways for this scenario = 56 * 220 = 12,320
Therefore, the total number of ways to form a committee of 6 members with at least 1 boy and 1 girl is:
6,336 + 13,860 + 12,320 = 32,516.
So, the correct answer is 32,516, not 896.
To find the number of ways a committee of 6 can be formed with at least 1 boy and 1 girl from a class of 12 boys and 8 girls, we can consider the two cases separately:
Case 1: If the committee consists of 1 boy and 5 girls:
The number of ways to select 1 boy from 12 is C(12, 1) = 12, and the number of ways to select 5 girls from 8 is C(8, 5) = 56. (Here, C(n, r) denotes the number of combinations of selecting r items from a set of n items.) Therefore, the total number of ways for this case is 12 * 56 = 672.
Case 2: If the committee consists of 2 or more boys and 4 or fewer girls:
For this case, we need to consider the number of possibilities for each combination of boys and girls. Let's analyze this step by step:
- Number of boys in the committee: 2, 3, 4, 5, or 6.
- Number of girls in the committee: 4, 3, 2, 1, or 0 (since the total committee size is 6).
To calculate the total number of possibilities for each combination, we can sum the products of C(12, r) (selecting r boys) and C(8, 6-r) (selecting 6-r girls) for each combination.
Summing all the possibilities for the combinations, we get:
C(12, 2) * C(8, 4) + C(12, 3) * C(8, 3) + C(12, 4) * C(8, 2) + C(12, 5) * C(8, 1) + C(12, 6) * C(8, 0) = 12 * 70 + 220 * 56 + 495 * 28 + 792 * 8 + 924 * 1 = 840 + 12320 + 13860 + 6336 + 924 = 33780.
Finally, we add the results from both cases:
Total number of ways = 672 + 33,780 = 34,452.
Hence, the correct answer is 34,452 ways.