There are 6 boys and 8 girls in a group of students. Calculate the number of ways that a committee of 3 students which has at least 1 girl can be selected from the group.

Number of ways to choose 3 students from 14 = C(14,3) = 14!/((14-3)!3!)

Number of ways to choose 3 students from 8 boys = C(8,3) = 8!/((8-3)!3!)

Number of ways to choose 3 students with at least one girl
= C(14,3)-C(8,3)

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To calculate the number of ways a committee of 3 students can be selected with at least 1 girl, we can use the principle of counting and apply combinations.

First, let's consider the case where all 3 students are girls. We can select 3 girls out of the 8 available in 8C3 ways, which is equal to:

8C3 = 8! / (3! * (8-3)!) = 8! / (3! * 5!) = (8 * 7 * 6) / (3 * 2 * 1) = 56

Now, let's consider the cases where exactly 2 girls are selected. We can select 2 girls out of the 8 available in 8C2 ways, and choose 1 boy out of the 6 available in 6C1 ways. This gives us a total of:

8C2 * 6C1 = (8! / (2! * (8-2)!) ) * (6! / (1! * (6-1)!)) = 28 * 6 = 168

Finally, let's consider the case where exactly 1 girl is selected. We can select 1 girl out of the 8 available in 8C1 ways, and choose 2 boys out of the 6 available in 6C2 ways. This gives us:

8C1 * 6C2 = (8! / (1! * (8-1)!) ) * (6! / (2! * (6-2)!)) = 8 * 15 = 120

There are no cases where all 3 students are boys since we require at least 1 girl to be selected.

To calculate the total number of ways that a committee of 3 students with at least 1 girl can be selected, we can sum up the three cases:

Total ways = 56 + 168 + 120 = 344

Therefore, there are 344 ways that a committee of 3 students with at least 1 girl can be selected from the group.