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Q2(ii):
Determine $n$ if (ii) ${}^{2n}C_3 : {}^nC_3 = 11 : 1$
Solution :
Step 1: Definition of the Combinatorial Formula
To solve for $n$, we utilize the standard formula for combinations, which defines the number of ways to select $r$ objects from a set of $n$ objects [Per the definition of $nCr$]:
${}^nC_r = \frac{n!}{r!(n-r)!}$
Step 2: Expanding the Given Ratio
Given the equation ${}^{2n}C_3 : {}^nC_3 = 11 : 1$, we write this as a fractional expression:
$\frac{{}^{2n}C_3}{{}^nC_3} = \frac{11}{1}$
Applying the factorial formula to both terms:
$\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 11$
Step 3: Algebraic Simplification
Cancel the constant $3!$ from both numerator and denominator [By the properties of algebraic fractions]:
$\frac{(2n)!}{(2n-3)!} \times \frac{(n-3)!}{n!} = 11$
Expand the factorials to eliminate $(2n-3)!$ and $(n-3)!$:
$\frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 11$
$\frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} = 11$
Step 4: Solving for $n$
Factor out common terms from the numerator:
$\frac{2n(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 11$
Cancel $n$ and $(n-1)$ from the numerator and denominator [Condition: $n > 3$ as per the definition of $nC_r$]:
$\frac{4(2n-1)}{n-2} = 11$
Multiply to solve the linear equation:
$4(2n - 1) = 11(n - 2)$
$8n - 4 = 11n - 22$
$22 - 4 = 11n - 8n$
$18 = 3n$
$n = 6$
Step 5: Verification of Geometric/Combinatorial Constraints
Final Solution: $n = 6$
More Questions from Class 11 Mathematics Permutations and Combinations EXERCISE 6.4
- Q1: If ${}^nC_8 = {}^nC_2$, find ${}^nC_2$.
- Q2(i): Determine $n$ if (i) ${}^{2n}C_3 : {}^nC_3 = 12 : 1$
- 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
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