Question

If ¹¹P_r=7920 and ¹¹C_r = 330, then the value of r is equal to

• A. 2
• B. 3
• C. 4
• D. 5
• E. 6

Answer 1

The Correct Option is C

We are given two formulas:

\[ 11P_r = 7920 \quad \text{(Permutation)} \] \[ 11C_r = 330 \quad \text{(Combination)} \]

Step 1: Use the formula for permutation

The formula for permutation is: \[ ^nP_r = \frac{n!}{(n - r)!} \] For \( 11P_r \), we have: \[ 11P_r = \frac{11!}{(11 - r)!} \] Substitute \( 11P_r = 7920 \): \[ \frac{11!}{(11 - r)!} = 7920 \] Now calculate \( 11! \): \[ 11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 39916800 \] Thus, the equation becomes: \[ \frac{39916800}{(11 - r)!} = 7920 \] Now, simplify: \[ (11 - r)! = \frac{39916800}{7920} = 5040 \]

Step 2: Find \( (11 - r)! = 5040 \)

Now, check the factorial values to find \( (11 - r) \): \[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \] So, \( 11 - r = 7 \), hence \( r = 4 \).

Step 3: Verify using the combination formula

We also have the combination formula: \[ ^nC_r = \frac{n!}{r!(n - r)!} \] Substitute \( 11C_r = 330 \) and \( r = 4 \): \[ 11C_4 = \frac{11!}{4!(11 - 4)!} = \frac{11!}{4!7!} \] Now calculate: \[ 4! = 4 \times 3 \times 2 \times 1 = 24 \] \[ 11C_4 = \frac{39916800}{24 \times 5040} = \frac{39916800}{120960} = 330 \] This confirms that \( r = 4 \).

The correct option is (C) : 4

Answer 2

We are given that ¹¹P_r = 7920 and ¹¹C_r = 330. We want to find the value of r.

We know that ⁿP_r = \(\frac{n!}{(n-r)!}\) and ⁿC_r = \(\frac{n!}{r!(n-r)!}\).

Therefore, ¹¹P_r = \(\frac{11!}{(11-r)!}\) = 7920 and ¹¹C_r = \(\frac{11!}{r!(11-r)!}\) = 330.

We can also write ¹¹P_r = r! × ¹¹C_r.

Plugging in the given values, we have:

7920 = r! × 330

r! = \(\frac{7920}{330} = 24\)

Now we need to find the value of r such that r! = 24. We know that:

• 1! = 1
• 2! = 2 × 1 = 2
• 3! = 3 × 2 × 1 = 6
• 4! = 4 × 3 × 2 × 1 = 24
• 5! = 5 × 4 × 3 × 2 × 1 = 120

Since 4! = 24, we have r = 4.

Therefore, the value of r is equal to 4.