Skip to content

Punjab Board Class 11 Mathematics

Exercise 7.1 Solutions — Permutations and Combinations

Complete worked solutions for Exercise 7.1 from Chapter 7 (Permutations and Combinations), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 7.1 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 7: Permutations and Combinations.

Question 1: Evaluate each of the following:

Har part mein bara factorial khol kar chhote factorial se cancel karenge. Is sawal mein hum factorial ki definition \(n! = n(n-1)!\) use karenge.

(i) \(\frac{10!}{0!\,8!}\)

\[ \frac{10!}{0! \cdot 8!} = \frac{10 \cdot 9 \cdot 8!}{1 \cdot 8!} = 10 \cdot 9 = 90 \]

(ii) \(\frac{12!}{3!(12-3)!}\)

\[ \frac{12!}{3! \cdot 9!} = \frac{12 \cdot 11 \cdot 10 \cdot 9!}{6 \cdot 9!} = \frac{1320}{6} = 220 \]

(iii) \(\frac{1440}{3!4!} + \frac{2400}{5!2!}\)

\[ \frac{1440}{6 \cdot 24} + \frac{2400}{120 \cdot 2} = \frac{1440}{144} + \frac{2400}{240} = 10 + 10 = 20 \]

(iv) \(\frac{(n+2)!}{(n+1)!}\)

\[ \frac{(n+2)!}{(n+1)!} = \frac{(n+2)(n+1)!}{(n+1)!} = n + 2 \]
Answer: (i) \(90\), (ii) \(220\), (iii) \(20\), (iv) \(n+2\)

Question 2: Write each of the following in the factorial form:

Product ko factorial mein badalne ke liye missing tail se multiply aur divide kar dete hain, taake poora factorial ban jaye.

(i) \(n^3 - n\)

\[ n^3 - n = n(n^2 - 1) = n(n-1)(n+1) = \frac{(n+1)!}{(n-2)!} \]

(ii) \(n(n-1)(n-2)\cdots(n-r+1)\)

\[ n(n-1)(n-2)\cdots(n-r+1) = \frac{n(n-1)\cdots(n-r+1)(n-r)!}{(n-r)!} = \frac{n!}{(n-r)!} \]
Answer: (i) \(\frac{(n+1)!}{(n-2)!}\), (ii) \(\frac{n!}{(n-r)!}\)

Question 3: Find \(n\), if \((n+4)! = 3024 \cdot n!\).

Dono taraf \(n!\) se divide kar do — bachne wala product \((n+4)(n+3)(n+2)(n+1)\) hoga, jise 3024 ke consecutive factors se match karenge.

\[ \frac{(n+4)!}{n!} = 3024 \]
\[ (n+4)(n+3)(n+2)(n+1) = 3024 \]
\[ 9 \cdot 8 \cdot 7 \cdot 6 = 3024 \implies n+1 = 6 \implies n = 5 \]
Answer: \(n = 5\)

Question 4: If \(\frac{1}{7!} + \frac{1}{8!} = \frac{x}{9!}\), find \(x\).

Poori equation ko \(9!\) se multiply kar do — factorials cancel ho jayenge aur seedha \(x\) nikal aayega.

\[ \frac{9!}{7!} + \frac{9!}{8!} = x \]
\[ (9 \cdot 8) + 9 = x \implies x = 72 + 9 = 81 \]
Answer: \(x = 81\)

Question 5: Prove that: \(\frac{(2n+1)!}{n!} = [1 \cdot 3 \cdot 5 \cdots (2n-1)(2n+1)]2^n\)

\((2n+1)!\) mein odd aur even factors alag kar lo. Even factors mein se \(2^n\) common nikalne par \(n!\) bachega jo denominator se cancel ho jayega.

\[ \text{LHS} = \frac{1 \cdot 2 \cdot 3 \cdots (2n+1)}{n!} \]
\[ = \frac{[1 \cdot 3 \cdot 5 \cdots (2n+1)] \cdot [2 \cdot 4 \cdot 6 \cdots 2n]}{n!} \]
\[ = \frac{[1 \cdot 3 \cdot 5 \cdots (2n+1)] \cdot 2^n(1 \cdot 2 \cdot 3 \cdots n)}{n!} \]
\[ = [1 \cdot 3 \cdot 5 \cdots (2n+1)] \cdot 2^n = \text{RHS} \]
Answer: Proved.

Question 6: Express as a single fraction: \(\frac{(n+2)!}{(r+2)!} + \frac{(n+1)!}{(r+1)!}\)

Dono fractions ka common denominator \((r+2)!(r+1)!\) lo, phir numerator mein \((n+1)!(r+1)!\) common nikaal kar simplify karo.

\[ = \frac{(n+2)!(r+1)! + (n+1)!(r+2)!}{(r+2)!(r+1)!} \]
\[ = \frac{(n+1)!(r+1)!\big[(n+2) + (r+2)\big]}{(r+2)!(r+1)!} = \frac{(n+1)!(n+r+4)}{(r+2)!} \]
Answer: \(\frac{(n+1)!(n+r+4)}{(r+2)!}\)

Question 7: There are four distinct coloured balls and four boxes of the same colours as those of the balls. Determine the number of possible ways the balls, one each in a box, can be placed such that no ball goes to the box of its own colour.

Yeh derangement ka sawal hai — koi bhi ball apne rang ke box mein na jaye. Derangement ka formula \(!n = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \cdots + \frac{(-1)^n}{n!}\right)\) hai.

\[ !4 = 4!\left(1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right) \]
\[ = 24\left(\frac{1}{2} - \frac{1}{6} + \frac{1}{24}\right) = 24\left(\frac{12 - 4 + 1}{24}\right) = 9 \]
Answer: \(9\) ways