Mathematics 11 (Punjab Board)
Exercise 7.4 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 7: Permutations and Combinations.
Question 1: Evaluate the following:
Jab \(r > \frac{n}{2}\) ho to \(^{n}C_{r} = ^{n}C_{n-r}\) use karo — calculation chhoti ho jayegi.
(i) \(^{50}C_{50} = 1\)
(ii) \(^{1000}C_{0} = 1\)
(iii) \(^{10}C_{7}\)
\[
^{10}C_{7} = ^{10}C_{3} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120
\]
(iv) \(^{20}C_{17}\)
\[
^{20}C_{17} = ^{20}C_{3} = \frac{20 \cdot 19 \cdot 18}{3 \cdot 2 \cdot 1} = 1140
\]
Answer: (i) \(1\), (ii) \(1\), (iii) \(120\), (iv) \(1140\)
Question 2: Find the value of \(n\) when:
Combination ka formula \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\) lagao, aur jahan \(^{n}P_{r}\) aur \(^{n}C_{r}\) dono hon wahan \(^{n}P_{r} = ^{n}C_{r} \times r!\) ka relation use karo.
(i) \(^{3n}C_{2} : ^{2}C_{2} = 15 : 1\)
\[
\frac{3n(3n-1)}{2} = 15 \implies 9n^2 - 3n - 30 = 0 \implies (3n+5)(n-2) = 0
\]
\[
n = 2 \quad (n positive integer hai)
\]
(ii) \(^{6}P_{r} = 120\) and \(^{6}C_{r} = 20\)
\[
120 = 20 \times r! \implies r! = 6 \implies r = 3
\]
Answer: (i) \(n = 2\), (ii) \(r = 3\)
Question 3: Find the values of \(n\) and \(r\), when:
Pehle \(^{n}P_{r} = ^{n}C_{r} \times r!\) se \(r\) nikalo, phir combination expand karke \(n\) nikalo. Ratio wale part mein consecutive combinations ka ratio \(\frac{^{n}C_{r}}{^{n+1}C_{r}} = \frac{n+1-r}{n+1}\) yaad rakho.
(i) \(^{n}C_{r} = 56\) and \(^{n}P_{r} = 336\)
\[
r! = \frac{336}{56} = 6 \implies r = 3
\]
\[
\frac{n(n-1)(n-2)}{6} = 56 \implies n(n-1)(n-2) = 336 = 8 \cdot 7 \cdot 6 \implies n = 8
\]
(ii) \(^{n-1}C_{r} : ^{n}C_{r} : ^{n+1}C_{r} = 1 : 3 : 7\)
\[
\frac{^{n-1}C_{r}}{^{n}C_{r}} = \frac{n-r}{n} = \frac{1}{3} \implies 2n = 3r
\]
\[
\frac{^{n}C_{r}}{^{n+1}C_{r}} = \frac{n+1-r}{n+1} = \frac{3}{7} \implies 4n - 7r = -4
\]
\[
4n - 7\left(\frac{2n}{3}\right) = -4 \implies -2n = -12 \implies n = 6, \; r = 4
\]
Answer: (i) \(n = 8, r = 3\), (ii) \(n = 6, r = 4\)
Question 4: Prove that:
LHS se start karo, \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\) rakho, common factor nikaal kar RHS tak simplify karo.
(i) \(^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}\)
\[
\text{LHS} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!}
\]
\[
= \frac{n!}{(r-1)!(n-r)!}\left[\frac{1}{r} + \frac{1}{n-r+1}\right] = \frac{n!}{(r-1)!(n-r)!} \cdot \frac{n+1}{r(n-r+1)}
\]
\[
= \frac{(n+1)!}{r!(n-r+1)!} = ^{n+1}C_{r} = \text{RHS}
\]
(ii) \(r \cdot ^{n}C_{r} = (n-r+1) \cdot ^{n}C_{r-1}\)
\[
\text{LHS} = r \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r)!}
\]
\[
\text{RHS} = (n-r+1) \cdot \frac{n!}{(r-1)!(n-r+1)!} = \frac{n!}{(r-1)!(n-r)!} = \text{LHS}
\]
Answer: Proved.
Question 5: Prove that the product of \(r\) consecutive integers is divisible by \(r!\).
\(r\) consecutive integers ka product \((n+r-1)!/(n-1)!\) hota hai — isay \(r!\) se multiply-divide karke combination ki shakal banao.
\[
n(n+1)\cdots(n+r-1) = \frac{(n+r-1)!}{(n-1)!} = r! \cdot \frac{(n+r-1)!}{r!(n-1)!} = r! \cdot ^{n+r-1}C_{r}
\]
Since \(^{n+r-1}C_{r}\) is an integer, the product equals \(r!\) times an integer — hence divisible by \(r!\).
Answer: Proved.
Question 6: In how many ways can five subjects be selected out of eight subjects to select a course programme?
Sirf selection hai, order nahi — combination \(^{8}C_{5}\).
\[
^{8}C_{5} = ^{8}C_{3} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56
\]
Answer: \(56\)
Question 7: Find the number of possible arrangements of the vowel letters from the English alphabet.
Vowels 5 hain (A, E, I, O, U) — sab ko arrange karna hai: \(5!\).
\[
5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120
\]
Answer: \(120\)
Question 8: In how many ways can 3 dishes of Desi foods and 2 dishes of Chinese foods be selected from 6 dishes of Desi foods and 8 dishes of Chinese foods?
Dono selections alag alag karo aur multiply karo: \(^{6}C_{3} \times ^{8}C_{2}\).
\[
^{6}C_{3} \times ^{8}C_{2} = 20 \times 28 = 560
\]
Answer: \(560\)
Question 9: From a standard deck of 52 playing cards, there are 26 black cards and 26 red cards. How many different ways can eight cards be selected if 3 are black and the remaining 5 are red?
Black cards mein se 3 aur red mein se 5 — dono combinations ko multiply karo.
\[
^{26}C_{3} \times ^{26}C_{5} = 2600 \times 65780 = 171028000
\]
Answer: \(171028000\)
Question 10: A bag contains 8 red balls and 7 green balls. Find the total number of possible ways in which five balls are selected such that: (i) 3 are red and 2 are green (ii) 1 is red and 4 are green (iii) 4 are red and 1 is green (iv) all are red.
Har condition mein red aur green ke selections alag nikaal kar multiply karo.
(i) 3 red and 2 green:
\[
^{8}C_{3} \times ^{7}C_{2} = 56 \times 21 = 1176
\]
(ii) 1 red and 4 green:
\[
^{8}C_{1} \times ^{7}C_{4} = 8 \times 35 = 280
\]
(iii) 4 red and 1 green:
\[
^{8}C_{4} \times ^{7}C_{1} = 70 \times 7 = 490
\]
(iv) All red:
\[
^{8}C_{5} = 56
\]
Answer: (i) \(1176\), (ii) \(280\), (iii) \(490\), (iv) \(56\)
Question 11: How many diagonals and triangles can be formed by joining the vertices of a polygon having 15 sides?
Diagonals = \(^{15}C_{2} - 15\); triangles = \(^{15}C_{3}\).
\[
\text{Diagonals} = \frac{15 \cdot 14}{2} - 15 = 105 - 15 = 90
\]
\[
\text{Triangles} = ^{15}C_{3} = \frac{15 \cdot 14 \cdot 13}{6} = 455
\]
Answer: \(90\) diagonals, \(455\) triangles
Question 12: Find the number of sides of a polygon if the number of its diagonals is 104.
Formula \(\frac{n(n-3)}{2} = 104\) lagao aur quadratic solve karo.
\[
\frac{n(n-3)}{2} = 104 \implies n^2 - 3n - 208 = 0 \implies (n-16)(n+13) = 0
\]
\[
n = 16 \quad (n positive hai)
\]
Answer: \(16\) sides
Question 13: How many triangles can be formed by joining 15 points, 6 of which lie on the same straight line?
Total triangles \(^{15}C_{3}\) mein se woh "triangles" minus karo jo collinear 6 points se bante (woh triangle nahi banate).
\[
^{15}C_{3} - ^{6}C_{3} = 455 - 20 = 435
\]
Answer: \(435\)
Question 14: The members of a club are 10 boys and 8 girls. In how many ways can a committee of 6 boys and 3 girls be formed?
\(^{10}C_{6} \times ^{8}C_{3}\) — dono selections ko multiply karo.
\[
^{10}C_{6} \times ^{8}C_{3} = 210 \times 56 = 11760
\]
Answer: \(11760\)
Question 15: How many committees of 7 members can be chosen from a group of 10 persons when each committee must include 2 particular persons?
2 khaas afraad to lazmi hain — baqi 5 members baqi 8 mein se chuno: \(^{8}C_{5}\).
\[
^{8}C_{5} = ^{8}C_{3} = 56
\]
Answer: \(56\)
Question 16: In how many ways can a cricket team of 11 players be selected out of 17 players? How many of them will include a particular player?
Total selections \(^{17}C_{11}\); khaas khiladi shamil ho to baqi 10, baqi 16 mein se: \(^{16}C_{10}\).
\[
^{17}C_{11} = ^{17}C_{6} = 12376, \qquad ^{16}C_{10} = ^{16}C_{6} = 8008
\]
Answer: \(12376\) teams, \(8008\) include the particular player
Question 17: There are 6 men and 8 women members of a club. How many committees of seven can be formed: (i) with 3 women (ii) with at most 3 women (iii) with at least 5 women?
Har case ko women ki tadad ke hisab se tor kar combinations multiply karo, phir add karo.
(i) With 3 women (and 4 men):
\[
^{8}C_{3} \times ^{6}C_{4} = 56 \times 15 = 840
\]
(ii) With at most 3 women:
\[
^{8}C_{1}\,^{6}C_{6} + ^{8}C_{2}\,^{6}C_{5} + ^{8}C_{3}\,^{6}C_{4} = 8 + 168 + 840 = 1016
\]
(iii) With at least 5 women:
\[
^{8}C_{5}\,^{6}C_{2} + ^{8}C_{6}\,^{6}C_{1} + ^{8}C_{7}\,^{6}C_{0} = 840 + 168 + 8 = 1016
\]
Answer: (i) \(840\), (ii) \(1016\), (iii) \(1016\)
Question 18: There are three sections in a question paper, each section has 3 questions. A student has to solve all 5 questions, choosing at least one question from each section. In how many ways can the student make his choice?
5 questions ki taqseem 3 sections mein sirf do tarah ho sakti hai: (3,1,1) ya (2,2,1). Dono cases nikaal kar add karo.
Case 1: distribution (3, 1, 1)
\[
^{3}C_{3}\,^{3}C_{1}\,^{3}C_{1} \times \frac{3!}{2!} = 9 \times 3 = 27
\]
Case 2: distribution (2, 2, 1)
\[
^{3}C_{2}\,^{3}C_{2}\,^{3}C_{1} \times \frac{3!}{2!} = 27 \times 3 = 81
\]
\[
\text{Total} = 27 + 81 = 108
\]
Answer: \(108\) ways
Question 19: Consider a cryptographic system that generates an 8-character password. Each character in the password can be either a lowercase letter (a–z) or a digit (0–9). How many passwords can be generated if each password must contain exactly 5 lowercase letters and 3 digits: (a) with repetition allowed (b) without repetition?
Pehle 8 mein se 5 positions letters ke liye chuno (\(^{8}C_{5}\)), phir unhein bharo. Repetition ho to powers, na ho to permutations.
(a) With repetition allowed:
\[
^{8}C_{5} \times 26^{5} \times 10^{3} = 56 \times 11881376 \times 1000 = 665357056000
\]
(b) Without repetition:
\[
^{8}C_{5} \times ^{26}P_{5} \times ^{10}P_{3} = 56 \times 7893600 \times 720 = 318269952000
\]
Answer: (a) \(665357056000\), (b) \(318269952000\)
Question 20: On Defense Day, Teacher I compiles a list of 10 distinct national songs, while Teacher II prepares a separate list of 10 different national songs (with no overlap between the two lists). The principal needs to select 3 songs from Teacher I's list, and 3 songs from Teacher II's list. Find the number of ways: (i) when the order of the songs is important (ii) when the order of the songs is not important.
Dono lists se selections alag alag hongi. Order important ho to permutation, warna combination — phir multiply.
(i) Order important:
\[
^{10}P_{3} \times ^{10}P_{3} = 720 \times 720 = 518400
\]
(ii) Order not important:
\[
^{10}C_{3} \times ^{10}C_{3} = 120 \times 120 = 14400
\]
Answer: (i) \(518400\), (ii) \(14400\)
