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This is what I got: 18C3 * 7C2 + 18C4 * 7C1 + 18C5 * 7C0 = 47,124

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Q: How many different teams of 5 children can be chosen from a group of 18 girls and 7 boys if each team must have at least two girls on it?

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The probability of choosing 2 girls at random from group of 25 students of which10 are girls and 15 are boys is:P( 2 girls) = (10/25)∙(9/24) = 3/20 = 0.15 = 15%

let boys = B and girls = GB +G =120;2G = B;3G = 120;G = 40 and B = 80

I figured out the answer: 18+7 = 25, So 25C5 = 53,130 Teams. This assumes there are no sex restrictions on the teams (for example, "teams must include at least one boy and at least one girl"). That would be a lot more complicated, and if that's a factor you'd need to specify exactly what the restrictions are.

If there are four times as many boys as girls, then the ratio of boys to girls is 4:1. To find the number of girls, divide the total number of children by the sum of the parts of the ratio, which is 4+1. So, there would be 30/5 = 6 girls in the group.

40 x 39 x 38 x 37 = 2193360

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The probability of choosing 2 girls at random from group of 25 students of which10 are girls and 15 are boys is:P( 2 girls) = (10/25)∙(9/24) = 3/20 = 0.15 = 15%

This means that for every girl there are two boys - so in a group of three, two will be boys and one a girl 45 ÷ 3 = 15 So, there are 15 girls and 30 boys.

This means that for every girl there are two boys - so in a group of three, two will be boys and one a girl 45 ÷ 3 = 15 So, there are 15 girls and 30 boys

It should be--- girls' group.

C = childrenB = boysG = girlsB = 3G3G + G = C4G = 60G = 15B = 45