Question

For a sports event of class V students, a packet of 7 caps is to be prepared from 9 green colored caps and 4 red caps. In how many ways can this be done when the packet contains at most 3 red caps?

• A. 1588 ways
• B. 1632 ways
• C. 1748 ways
• D. 1684 ways

Hint:

When calculating combinations with restrictions, break the problem into cases and compute the number of ways for each scenario. Then sum the results to find the total number of ways.

Answer 1

The Correct Option is B

Step 1: Understanding the Problem.
We are tasked with selecting a total of 7 caps, consisting of green and red caps. The packet can contain at most 3 red caps. We need to find how many ways this can be done, ensuring the number of red caps is no more than 3.
Step 2: Possible Combinations.
The total number of red caps can be 0, 1, 2, or 3. We will calculate the number of ways to choose green and red caps for each case. - **Case 1: 0 red caps** If there are 0 red caps, we select all 7 caps from the 9 green caps. The number of ways to do this is: \[ \binom{9}{7} = 36 \] - **Case 2: 1 red cap** If there is 1 red cap, we select 6 green caps from the 9 green caps. The number of ways to do this is: \[ \binom{9}{6} \times \binom{4}{1} = 84 \times 4 = 336 \] - **Case 3: 2 red caps** If there are 2 red caps, we select 5 green caps from the 9 green caps. The number of ways to do this is: \[ \binom{9}{5} \times \binom{4}{2} = 126 \times 6 = 756 \] - **Case 4: 3 red caps** If there are 3 red caps, we select 4 green caps from the 9 green caps. The number of ways to do this is: \[ \binom{9}{4} \times \binom{4}{3} = 126 \times 4 = 504 \] Step 3: Total Number of Ways.
Now, we sum the number of ways for all cases: \[ 36 + 336 + 756 + 504 = 1632 \] Step 4: Conclusion.
The total number of ways to select the caps is 1632, which corresponds to option (2).