Correct Answer: Option B

Explanation

Given: \(\frac{^6C_r}{^6P_r} = \frac{1}{6}\)

Recall the formulas:  
\(^nC_r = \frac{n!}{r!(n-r)!}, \quad ^nP_r = \frac{n!}{r!}\)

Substitute \( n = 6 \):

\(\frac{^6C_r}{^6P_r} = \frac{\dfrac{6!}{r!(6-r)!}}{\dfrac{6!}{r!}} = \frac{6!}{r!(6-r)!} \times \frac{r!}{6!} = \frac{1}{(6-r)!}\)

So the equation simplifies to:
\(\frac{1}{(6-r)!} = \frac{1}{6}\)

Therefore: (6-r)! = 6

We know that 3! = 6, so:
6 - r = 3 \(\implies\) r = 3

• a compund microscope by lens of focal length 1cm and 5cm.an object placed 1.2cm from the objective lens if a... Answer this

  1 Answers   ·   Physics   ·   1 day ago
• if 5^(2x)=125 find x? Answer this

  1 Answers   ·   Mathematics   ·   17 hours ago
• state 5 reasons why promotion of unity and integration contributes to national development? Answer this

  3 Answers   ·   1 day ago