$ ext{ If } {}^{15}C_{3r} = {}^{15}C_{r+3}, ext{ then 'r' is equal to: }$

Question

$	ext{ If } {}^{15}C_{3r} = {}^{15}C_{r+3}, 	ext{ then 'r' is equal to: }$

Accepted Answer

To solve the problem, we need to understand the properties of combinations. The given equation is:

$$ {}^{15}C_{3r} = {}^{15}C_{r+3} $$

The formula for combinations is:

$$ {}^nC_k = \frac{n!}{k!(n-k)!} $$

Applying this formula to both sides of the equation, we have:

$$ \frac{15!}{(3r)!(15-3r)!} = \frac{15!}{(r+3)!(15-(r+3))!} $$

Simplifying the right side:

$$ \frac{15!}{(r+3)!(12-r)!} $$

Since the factorial terms are equal, we can equate the denominators:

$$ (3r)!(15-3r)! = (r+3)!(12-r)! $$

To solve for $ r $, we need to find a value that satisfies this equation. One useful property of combinations is that:

$$ {}^nC_k = {}^nC_{n-k} $$

This means:

$$ {}^{15}C_{3r} = {}^{15}C_{15-3r} $$

Given that:

$$ {}^{15}C_{3r} = {}^{15}C_{r+3} $$

We can equate:

$$ 15 - 3r = r + 3 $$

Solving for $ r $:

$$ 15 - 3r = r + 3 $$

$$ 15 - 3 = 4r $$

$$ 12 = 4r $$

$$ r = 3 $$

Thus, $ r = 3 $ satisfies the equation, confirming that Option 3 is correct.

**Explanation for Incorrect Options:**

1. **Option 1 (r = 5):**
   - If $ r = 5 $, then $ 3r = 15 $ and $ r+3 = 8 $.
   - The equation becomes $ {}^{15}C_{15} = {}^{15}C_8 $, which is not true because $ {}^{15}C_{15} = 1 $ and $ {}^{15}C_8 $ is not equal to 1.

2. **Option 2 (r = 4):**
   - If $ r = 4 $, then $ 3r = 12 $ and $ r+3 = 7 $.
   - The equation becomes $ {}^{15}C_{12} = {}^{15}C_7 $, which is not true because $ {}^{15}C_{12} \neq {}^{15}C_7 $.

3. **Option 4 (r = 2):**
   - If $ r = 2 $, then $ 3r = 6 $ and $ r+3 = 5 $.
   - The equation becomes $ {}^{15}C_6 = {}^{15}C_5 $, which is not true because $ {}^{15}C_6 \neq {}^{15}C_5 $.

Therefore, the only value of $ r $ that satisfies the given equation is $ r = 3 $, making Option 3 the correct answer.