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Q2(i):
Determine $n$ if (i) ${}^{2n}C_3 : {}^nC_3 = 12 : 1$
Solution :
Step 1: Formulaic Representation and Given Constraints
The definition of the combination formula is given by ${}^{n}C_r = \frac{n!}{r!(n-r)!}$, where $n \geq r \geq 0$. We are provided with the ratio:
$\frac{{}^{2n}C_3}{{}^{n}C_3} = \frac{12}{1}$
Step 2: Expansion of the Combination Terms
Applying the formula to both components of the ratio:
${}^{2n}C_3 = \frac{(2n)!}{3!(2n-3)!} = \frac{(2n)(2n-1)(2n-2)(2n-3)!}{6(2n-3)!} = \frac{2n(2n-1)(2n-2)}{6}$
${}^{n}C_3 = \frac{n!}{3!(n-3)!} = \frac{n(n-1)(n-2)(n-3)!}{6(n-3)!} = \frac{n(n-1)(n-2)}{6}$
Step 3: Algebraic Setup and Simplification
Substituting these expressions back into the ratio equation:
$\frac{\left[ \frac{2n(2n-1)(2n-2)}{6} \right]}{\left[ \frac{n(n-1)(n-2)}{6} \right]} = 12$
[Per the property of division of fractions, the denominators (6) cancel out]:
$\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 12$
Step 4: Factoring and Reduction
Observe that $(2n-2) = 2(n-1)$. Substitute this into the numerator:
$\frac{2n(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 12$
[By cancelling common factors $n$ and $(n-1)$ from the numerator and denominator, noting $n > 2$]:
$\frac{4(2n-1)}{(n-2)} = 12$
Step 5: Solving for $n$
Divide both sides by 4:
$\frac{2n-1}{n-2} = 3$
[By cross-multiplication]:
$2n - 1 = 3(n - 2)$
$2n - 1 = 3n - 6$
[Rearranging terms to solve for $n$]:
$6 - 1 = 3n - 2n$
$n = 5$
Verification Diagram
Final Solution: n = 5
More Questions from Class 11 Mathematics Permutations and Combinations EXERCISE 6.4
- Q1: If ${}^nC_8 = {}^nC_2$, find ${}^nC_2$.
- Q2(ii): Determine $n$ if (ii) ${}^{2n}C_3 : {}^nC_3 = 11 : 1$
- Q3: How many chords can be drawn through 21 points on a circle?
- Q4: In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
- Q5: Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
- Q6: Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
- Q7: In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
- Q8: A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
- Q9: In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
CBSE Solutions for Class 11 Mathematics Permutations and Combinations
Chapters in CBSE - Class 11 Mathematics
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