How many different combinations of 4 students can be selected from a group of 12 students for a university committee?

To determine how many different combinations of 4 students can be selected from a group of 12 students, we use the concept of combinations in combinatorics. Combinations are used when the order of selection does not matter.

The formula for combinations is given by:

C(n, r) = n! / (r! * (n – r)!)

Where:

  • C(n, r) is the number of combinations,
  • n is the total number of items (in this case, students),
  • r is the number of items to choose, and
  • ! denotes factorial, the product of all positive integers up to that number.

For this specific problem:

  • n = 12 (the total number of students)
  • r = 4 (the number of students to select)

Now, we can substitute the values into the formula:

C(12, 4) = 12! / (4! * (12 – 4)!)

This simplifies to:

C(12, 4) = 12! / (4! * 8!)

Next, we can expand the factorials:

12! = 12 × 11 × 10 × 9 × 8!

Notice that the 8! in the numerator and denominator will cancel each other out, giving us:

C(12, 4) = (12 × 11 × 10 × 9) / (4 × 3 × 2 × 1)

Calculating the numerator:

  • 12 × 11 = 132
  • 132 × 10 = 1320
  • 1320 × 9 = 11880

This gives us a numerator of 11880.

Now for the denominator:

  • 4 × 3 = 12
  • 12 × 2 = 24
  • 24 × 1 = 24

Now we divide the numerator by the denominator:

C(12, 4) = 11880 / 24 = 495

Therefore, there are 495 different combinations of 4 students that can be selected from a group of 12 students to serve on the university committee.

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