A committee of 4 students is to be selected from a group of 12 students. Each student can be selected at most once, and the order in which the students are chosen does not matter.Which of the following expressions gives the number of possible committees?[Fill in the blank]Single choice
A
a. [math: (124)]
B
b. [math: (12+4−14)]
C
c. [math: 124]
D
d. [math: 12!(12−4)!]

Question

A committee of 4 students is to be selected from a group of 12 students. Each student can be selected at most once, and the order in which the students are chosen does not matter.Which of the following expressions gives the number of possible committees?[Fill in the blank]Single choice

A

a. [math: (124)]

B

b. [math: (12+4−14)]

C

c. [math: 124]

D

d. [math: 12!(12−4)!]

BlackTom AI · Accepted Answer

Choosing 4 students from 12 without repetition and without regard to order is a combinations problem. The relevant formula is the binomial coefficient: C(n, k) = n! / (k!(n − k)!), which counts the number of k-element subsets from an n-element set.

Step-by-step reasoning:
1. Identify n and k: n = 12 students, k = 4 chosen.
2. Apply the combination formula: C(12, 4) = 12! / (4!·(12 − 4)!).
3. Compute intermediate factorial factors to simplify:
   - 12! / (8!·4!) = (12 × 11 × 10 × 9 × 8!) / (8! × 4!) = (12 × 11 × 10 × 9) / 4!.
   - 4! = 4 × 3 × 2 × 1 = 24.
   - Numerator: 12 × 11 × 10 × 9 = 11,880.
   - Division: 11,880 / 24 = 495.
So C(12,4) = 495 possible committees.

Why option (a) is correct:
Option (a) displays the combination (12 choose 4), exactly matching the situation of selecting 4 distinct students from 12 where order doesn’t matter and repetition is not allowed. The computed value confirms this expression.

Why the other options are incorrect:
- (b) (12 + 4 − 1 choose 4) = C(15,4) is the formula for combinations with repetition (stars-and-bars); it applies when members can be chosen more than once, which is not allowed here.
- (c) 12^4 counts ordered 4-tuples with repetition allowed (each of 4 positions independently chosen from 12), so it overcounts and allows repeats and order.
- (d) 12!/(12−4)! = P(12,4) = 11,880 counts permutations (ordered selections without repetition); it distinguishes order, which is not required for a committee.

Conclusion: The correct expression is C(12,4), i.e., option (a), which equals 495, matching the provided answer.

A committee of 4 students is to be selected from a group of 12 students. Each student can be selected at most once, and the order in which the students are chosen does not matter.Which of the following expressions gives the number of possible committees?[Fill in the blank]Single choice

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