Mathematics 11 (Punjab Board)
Exercise 7.3 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 7: Permutations and Combinations.
Question 1: How many arrangements of the letters of the following words, taken all together can be made?
Repeated letters hon to \(\frac{n!}{p!\,q!\cdots}\) ka formula lagao — har repeated group ke factorial se divide karo.
(i) PAKISTAN (8 distinct letters)
\[
8! = 40320
\]
(ii) CURRICULUM (10 letters: C twice, U thrice, R twice, I twice)
\[
\frac{10!}{2!\,3!\,2!\,2!} = \frac{3628800}{48} = 75600
\]
(iii) PROBABILITY (11 letters: B twice, I twice)
\[
\frac{11!}{2!\,2!} = \frac{39916800}{4} = 9979200
\]
Answer: (i) \(40320\), (ii) \(75600\), (iii) \(9979200\)
Question 2: How many permutations of the letters of the word "BANANA" can be made, if B must be the first letter in each arrangement?
B fix ho gaya — baqi 5 letters (A 3 dafa, N 2 dafa) arrange karo.
\[
\frac{5!}{3!\,2!} = \frac{120}{12} = 10
\]
Answer: \(10\)
Question 3: How many arrangements of the letters of the word TRIGONOMETRY can be made, if each arrangement begins with T and ends with Y?
T aur Y fix — darmiyan wale 10 letters arrange karo, jin mein R 2 dafa aur O 3 dafa hai.
\[
\frac{10!}{2!\,3!} = \frac{3628800}{12} = 302400
\]
Answer: \(302400\)
Question 4: Abdullah has a collection of 9 marbles consisting of 4 identical red marbles, 3 identical blue marbles and 2 identical green marbles. If he wants to arrange all of them in a straight row, how many distinct arrangements are possible?
Identical marbles — 9 positions mein 4 red, 3 blue, 2 green ki taqseem.
\[
\frac{9!}{4!\,3!\,2!} = \frac{362880}{24 \cdot 6 \cdot 2} = \frac{362880}{288} = 1260
\]
Answer: \(1260\)
Question 5: In how many different ways can the following persons sit around a round table?
Gol mez par \((n-1)!\) tareeqe hote hain — aik shakhs ko reference bana kar fix kar do.
(a) 8 persons
\[
(8-1)! = 7! = 5040
\]
(b) 7 persons
\[
(7-1)! = 6! = 720
\]
(c) 6 persons
\[
(6-1)! = 5! = 120
\]
Answer: (a) \(5040\), (b) \(720\), (c) \(120\)
Question 6: In how many ways can 5 couples sit around a round table if no two women are sitting together?
Pehle 5 mardon ko gol mez par bithao — unke darmiyan 5 gaps banenge, jin mein 5 aurton ko bithao.
\[
(5-1)! \times 5! = 24 \times 120 = 2880
\]
Answer: \(2880\)
Question 7: How many 6-digit numbers can be formed from the digits 7, 7, 8, 8, 9, 9?
6 digits mein har aik 2 dafa — \(\frac{6!}{2!\,2!\,2!}\). (Pehla digit zero nahi, isliye koi pabandi nahi.)
\[
\frac{6!}{2!\,2!\,2!} = \frac{720}{8} = 90
\]
Answer: \(90\)
Question 8: 15 members of a club form 4 committees of 3, 5, 4 and 3 members so that no member is a member of more than one committee. Find the number of ways the committees can be formed.
Pehli committee \(^{15}C_{3}\), phir bachon mein se \(^{12}C_{5}\), phir \(^{7}C_{4}\), aakhir mein \(^{3}C_{3}\) — sab ko multiply karo.
\[
^{15}C_{3} \times ^{12}C_{5} \times ^{7}C_{4} \times ^{3}C_{3} = 455 \times 792 \times 35 \times 1 = 12612600
\]
Answer: \(12612600\)
Question 9: The D.C.Os of 11 districts meet to discuss the law-and-order situation in their districts. In how many ways can they be seated at a round table, when two particular D.C.Os insist on sitting together?
Dono ko aik block bana lo — ab 10 units gol mez par, block ke andar 2! tartib.
\[
(10-1)! \times 2! = 9! \times 2 = 362880 \times 2 = 725760
\]
Answer: \(725760\)
Question 10: The Governor of the Punjab calls a meeting of 14 officers. In how many ways can they be seated at a round table?
Seedha circular permutation: \((14-1)!\).
Answer: \(6227020800\)
Question 11: Fatima invites 14 people for a dinner. There are 9 males and 5 females who are seated at two different tables. Guests of one sex sit at one round table and the guests of the other sex sit at the second table. Find the number of ways in which all guests can be seated.
Dono tables par alag alag circular arrangements — \((9-1)!\) aur \((5-1)!\) ko multiply karo.
\[
8! \times 4! = 40320 \times 24 = 967680
\]
Answer: \(967680\)
Question 12: Find the number of ways in which 5 men and 5 women can be seated at a round table in such a way that no two persons of the same sex sit together.
Alternate sexes hon to pehle 5 mardon ko bithao, phir unke darmiyan wali 5 gaps mein 5 aurton ko.
\[
(5-1)! \times 5! = 24 \times 120 = 2880
\]
Answer: \(2880\)
Question 13: In how many ways can 8 keys be arranged in a circular key ring?
Key ring ko ulta bhi kiya ja sakta hai — clockwise aur anticlockwise aik jaise, isliye \(\frac{(8-1)!}{2}\).
\[
\frac{(8-1)!}{2} = \frac{5040}{2} = 2520
\]
Answer: \(2520\)
Question 14: How many necklaces can be made from 10 beads of different colours?
Necklace flip ho sakta hai, isliye \(\frac{(10-1)!}{2}\).
\[
\frac{(10-1)!}{2} = \frac{362880}{2} = 181440
\]
Answer: \(181440\)
