Mathematics 11 (Punjab Board)
Exercise 7.2 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 7: Permutations and Combinations.
Question 1: Evaluate the following:
Permutation ka formula \(^{n}P_{r} = \frac{n!}{(n-r)!}\) lagao aur factorial khol kar cancel karo.
(i) \(^{10}P_{5}\)
\[
^{10}P_{5} = \frac{10!}{5!} = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 30240
\]
(ii) \(^{5}P_{2}\)
\[
^{5}P_{2} = \frac{5!}{3!} = 5 \cdot 4 = 20
\]
(iii) \(^{7}P_{7}\)
\[
^{7}P_{7} = \frac{7!}{0!} = 7! = 5040 \qquad (\text{since } 0! = 1)
\]
(iv) \(^{10}P_{7}\)
\[
^{10}P_{7} = \frac{10!}{3!} = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 = 604800
\]
Answer: (i) \(30240\), (ii) \(20\), (iii) \(5040\), (iv) \(604800\)
Question 2: Find the value of \(n\) when:
Permutation ko expand karke consecutive product banao, phir number ko consecutive factors mein tor kar \(n\) pehchano.
(i) \(^{n}P_{3} = 504\)
\[
n(n-1)(n-2) = 504 = 9 \cdot 8 \cdot 7 \implies n = 9
\]
(ii) \(^{15}P_{n} = 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11\)
\[
\frac{15!}{(15-n)!} = \frac{15!}{10!} \implies 15 - n = 10 \implies n = 5
\]
(iii) \(^{n}P_{5} : ^{n-1}P_{2} = 540 : 1\)
\[
\frac{^{n}P_{5}}{^{n-1}P_{2}} = \frac{n(n-1)(n-2)(n-3)(n-4)}{(n-1)(n-2)} = n(n-3)(n-4) = 540
\]
Common mistake: The equation \(n(n-3)(n-4) = 540\) has no integer solution (\(10 \cdot 7 \cdot 6 = 420\) and \(11 \cdot 8 \cdot 7 = 616\)). The printed ratio 540 is a typo — with ratio \(420:1\) the answer is \(n = 10\).
Answer: (i) \(n = 9\), (ii) \(n = 5\), (iii) no integer \(n\) exists (see note; \(n = 10\) fits ratio \(420:1\))
Question 3: Prove from the first principle that:
First principle matlab definition \(^{n}P_{r} = \frac{n!}{(n-r)!}\) se start karo aur dono taraf ko isi form mein likh kar match karo.
(i) \(^{n}P_{r} = n \cdot ^{n-1}P_{r-1}\)
\[
\text{RHS} = n \cdot \frac{(n-1)!}{(n-1-(r-1))!} = n \cdot \frac{(n-1)!}{(n-r)!} = \frac{n!}{(n-r)!} = ^{n}P_{r} = \text{LHS}
\]
(ii) \(^{n}P_{r} = ^{n-1}P_{r} + r \cdot ^{n-1}P_{r-1}\)
\[
\text{RHS} = \frac{(n-1)!}{(n-1-r)!} + r \cdot \frac{(n-1)!}{(n-r)!}
\]
\[
= \frac{(n-1)!}{(n-r-1)!}\left[1 + \frac{r}{n-r}\right] = \frac{(n-1)!}{(n-r-1)!} \cdot \frac{n}{n-r} = \frac{n!}{(n-r)!} = ^{n}P_{r} = \text{LHS}
\]
Answer: Proved.
Question 4: How many words can be formed from the letters of the following words using all letters when no letter is to be repeated:
Sab letters different hain aur sab use hone hain — yeh \(n!\) arrangements honge.
(i) PYTHON
\[
6! = 720
\]
(ii) NETWORK
\[
7! = 5040
\]
(iii) COMPUTER
\[
8! = 40320
\]
Answer: (i) \(720\), (ii) \(5040\), (iii) \(40320\)
Question 5: How many signals can be given by 6 flags of different colours, using 2 flags at a time?
Signal mein flags ki tartib matter karti hai (upar-neeche), isliye yeh permutation hai: \(^{6}P_{2}\).
\[
^{6}P_{2} = 6 \cdot 5 = 30
\]
Answer: \(30\)
Question 6: How many signals can be given by 5 flags of different colours, when any number of flags are used at a time.
1 flag, 2 flags, ... 5 flags — har case ki permutations nikaal kar add kar do (addition principle).
\[
^{5}P_{1} + ^{5}P_{2} + ^{5}P_{3} + ^{5}P_{4} + ^{5}P_{5} = 5 + 20 + 60 + 120 + 120 = 325
\]
Answer: \(325\)
Question 7: How many 4 digit numbers can be formed, with distinct digits, with each digit odd?
Odd digits 1, 3, 5, 7, 9 — 5 mein se 4 ko order mein chunna hai: \(^{5}P_{4}\).
\[
^{5}P_{4} = 5 \cdot 4 \cdot 3 \cdot 2 = 120
\]
Answer: \(120\)
Question 8: How many numbers between 100 and 1000 can be formed by using the digits 0, 1, 2, 3, 4, 5 without repetition? How many of them are divisible by 5?
3-digit numbers — pehla digit zero nahi ho sakta. 5 se divisible hone ke liye aakhri digit 0 ya 5 hona chahiye.
Total 3-digit numbers:
\[
5 \times ^{5}P_{2} = 5 \times 20 = 100
\]
Divisible by 5:
\[
\text{Last digit } 0: 5 \times 4 = 20, \qquad \text{Last digit } 5: 4 \times 4 = 16
\]
\[
20 + 16 = 36
\]
Answer: \(100\) numbers, of which \(36\) are divisible by 5
Question 9: Find the numbers greater than 35000 that can be formed from the digits 1, 2, 3, 4, 5, 6, without repeating any digit.
5-digit numbers — pehle digit ke cases banao: 4, 5, 6 ho to sab chalen ge; 3 ho to doosra digit 5 ya 6 hona chahiye.
Case 1: first digit 4, 5 or 6
\[
3 \times ^{5}P_{4} = 3 \times 120 = 360
\]
Case 2: first digit 3, second digit 5 or 6
\[
1 \times 2 \times ^{4}P_{3} = 2 \times 24 = 48
\]
\[
\text{Total} = 360 + 48 = 408
\]
Answer: \(408\)
Question 10: Find the number of 5-digit numbers that can be formed from the digits 1, 2, 4, 6, 8 (when no digit is repeated), but (i) the digits 2 and 8 are next to each other; (ii) the digits 2 and 8 are not next to each other.
"Sath hon" ke liye 2 aur 8 ko aik block bana lo (andar 2! tartib). "Sath na hon" ke liye total mein se sath wale minus kar do.
(i) 2 and 8 next to each other:
\[
4! \times 2! = 24 \times 2 = 48
\]
(ii) 2 and 8 not next to each other:
\[
5! - 48 = 120 - 48 = 72
\]
Answer: (i) \(48\), (ii) \(72\)
Question 11: How many 6-digit numbers can be formed, without repeating any digit from the digits 0, 1, 2, 3, 4, 5? In how many of them will 0 be at the tens place?
Pehla digit zero nahi ho sakta — is restriction ko sab se pehle handle karo.
Total 6-digit numbers:
\[
5 \times ^{5}P_{5} = 5 \times 120 = 600
\]
0 at tens place:
\[
5 \times ^{4}P_{4} = 5 \times 24 = 120
\]
Answer: \(600\) numbers, \(120\) with 0 at the tens place
Question 12: How many 5-digit multiples of 5 can be formed from the digits 2, 3, 5, 7, 9, when no digit is repeated.
5 se divisible hone ke liye aakhri digit 5 hona lazmi hai (diye gaye digits mein sirf 5 kaam karega). Baqi 4 positions par baqi 4 digits arrange karo.
\[
^{4}P_{4} \times 1 = 24
\]
Answer: \(24\)
Question 13: In how many ways can 8 different books including 2 on English be arranged on a shelf in such a way that the English books are never together?
"Kabhi sath na hon" = total arrangements minus woh jin mein dono sath hon.
\[
8! - 7! \times 2! = 40320 - 10080 = 30240
\]
Answer: \(30240\)
Question 14: Find the number of arrangements of 3 different books on English and 5 different books on Urdu for placing them on a shelf such that the books on the same subject are together.
Har subject ka aik block banao: 2 blocks ko \(2!\) tareeqon se, English block ke andar \(3!\) se, Urdu block ke andar \(5!\) se arrange karo.
\[
2! \times 3! \times 5! = 2 \times 6 \times 120 = 1440
\]
Answer: \(1440\)
Question 15: In how many ways can 5 boys and 4 girls be seated on a bench so that the girls and the boys occupy alternate seats?
5 larke aur 4 larkiyan alternate hon to tartib sirf aik ho sakti hai: B G B G B G B G B. Larkon ko \(5!\) aur larkiyon ko \(4!\) tareeqon se bithao.
\[
5! \times 4! = 120 \times 24 = 2880
\]
Answer: \(2880\)
