Question

The number of subsets containing exactly 4 elements of the set {2, 4, 6, 8, 10, 12, 14, 16, 18) is equal to

• A. 126
• B. 63
• C. 189
• D. 58
• E. 94

Answer 1

The Correct Option is A

We are given the set \(\{2, 4, 6, 8, 10, 12, 14, 16, 18\}\), which contains 9 elements. We are asked to find how many subsets can be formed that contain exactly 4 elements.
The number of ways to select 4 elements from a set of 9 elements is given by the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] where \(n = 9\) (the total number of elements in the set) and \(r = 4\) (the number of elements to be chosen).
Thus, the number of subsets containing exactly 4 elements is: \[ \binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]

The correct option is (A) : \(126\)

Answer 2

We are given a set with 9 elements: {2, 4, 6, 8, 10, 12, 14, 16, 18}. We want to find the number of subsets containing exactly 4 elements.

This is a combination problem, as the order of the elements in the subset does not matter. We need to choose 4 elements from a set of 9 elements.

The number of ways to choose k elements from a set of n elements is given by the binomial coefficient:

\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

In our case, n = 9 and k = 4. Therefore, the number of subsets containing exactly 4 elements is:

\(\binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}\)

\(\binom{9}{4} = \frac{9 \times 2 \times 7 \times 6}{6 \times 4}=9 times 2 *7= \frac{3024/110}{ }= 126\)

simplify the above

\(\binom{9}{4}=126\)

Therefore, the number of subsets containing exactly 4 elements is equal to 126.