Question

Let \( A = \{ 1, 3, 5, 7, \dots, 21 \} \). The number of ways 4 numbers, containing always 11, can be selected from the set A is equal to

• A. 120
• B. 160
• C. 240
• D. 260
• E. 320

Hint:

When selecting a fixed number and then choosing the remaining elements, use combinations to calculate the number of ways.

Answer 1

The Correct Option is A

We are asked to select 4 numbers from the set \( A = \{1, 3, 5, 7, \dots, 21\} \) with the condition that 11 must always be included. 
The set \( A \) contains 11 odd numbers: \[ A = \{ 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 \} \] We are selecting 4 numbers in total, and one of them is fixed as 11. 
This leaves us with 3 more numbers to choose from the remaining 10 numbers: \[ \{ 1, 3, 5, 7, 9, 13, 15, 17, 19, 21 \} \] The number of ways to select 3 numbers from these 10 is given by the combination formula \( \binom{10}{3} \), which is: \[ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] Thus, the correct answer is (A).