Question

In how many ways can 8 identical pens be distributed among Aal, Bal, and Cal so that Aal gets at least 1 pen, Bal gets at least 2 pens, and Cal gets at least 3 pens?

• A. 4
• B. 3
• C. 2
• D. 6
  (E) 5
• E. 5

Hint:

To handle "at least" conditions with identical items, just subtract the total required items from the initial total. Then solve the problem for the remaining items as a standard whole-number distribution.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When there are minimum requirements for each person, we first satisfy those requirements by "giving away" the necessary pens. We then distribute the remaining pens using the standard formula.
Step 2: Detailed Explanation:
1. Total pens = 8.
2. Minimum requirements: - Aal: 1 pen - Bal: 2 pens - Cal: 3 pens 3. Total pens already assigned = $1 + 2 + 3 = 6$ pens.
4. Remaining pens to distribute = $8 - 6 = 2$ pens.
5. Now, we distribute these 2 identical pens among 3 distinct people with no further restrictions ($n=2, r=3$): \[ ^{2+3-1}C_{3-1} = \, ^4C_2 \] 6. Calculate the value: \[ ^4C_2 = \frac{4 \times 3}{2 \times 1} = 6. \]
Step 3: Final Answer:
The pens can be distributed in 6 ways.