Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 7 Permutations and Combinations Ex 7.1 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.1
Question 1.
How many 3-digit numbers can be formed from the digits 1,2,3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
Answer.
(i) There are five digits, 1, 2, 3, 4 and 5.
Every digits can be selected any number of times.
Hence, we can select first digit 5 times.
The second digits 5 times and third digit 5 times.
The number of ways in which the selection of three digits made = 5 × 5 × 5= 125.
(ii) Under the restriction, first digit can be selected in 5 ways.
After the selection of first digit four digits are left second digit can be selected in 4 ways and third digits can be selected in 3 ways.
Question 2.
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can he repeated?
Answer.
There will be as many ways as there are ways,of filling 3 vacant places 
in succession by the given six digits.
In this case, the units place can be filled by 2 or 4 or 6 only i.e., the units place can be filled in 3 ways.
The tens place can be filled by any of the 6 digits in 6 different ways and also the hundreds place can be filled by any of the 6 digits in 6 different ways, as the digits can be repeated.
Therefore, by multiplication principle, the required number of three digit even numbers is 3 × 6 × 6 = 108.
Question 3.
How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Answer.
First letter of the code word can be selected in 10 ways. After the selection of the first letter, we have 9 letters.
Hence, the second letter can be selected in 9 ways. According to FPC, first two letters can be selected in 10 × 9 ways.
Similarly, third letter can be selected in 8 ways and the fourth letter can be selected in 7 ways.
According to FPC (Fundamental Principle of Counting), number of ways of selecting four letters out of 10 letters of English alphabet is = 10 × 9 × 8 × 7 = 5040.
Question 4.
How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Answer.
It is given that the 5-digit telephone numbers always start with 67.
Therefore, there will be as many phone numbers as there are ways of filling 3 vacant places 
by the digits 0 – 9, keeping in mind that the digits cannot be repeated.
The units place can be filled by any of the digits from 0 – 9, except digits 6 and 7.
Therefore, the units place can be filled in 8 different ways following which, the tens place can be filled in by any of the remaining 7 digits in 7 different ways, and the hundreds place can be filled in by any of the remaining 6 digits in 6 different ways.
Therefore, by multiplication principle, the required number of ways in which 5-digit telephone numbers can be constructed is 8 × 7 × 6 = 336.
Question 5.
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Answer.
When a coin is tossed once, the number of outcomes is 2 (Head and tail) i.e., in each throw, the number of ways of showing a different face is 2.
Thus, by multiplication principle, the required number of possible outcomes is 2 × 2 × 2 = 8.
Question 6.
Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Answer.
E0ach signal requires the use of 2 flags.
There will be as many flags as there are ways of filling in 2 vacant places 
in succession by the given 5 flags of different colours.
The upper vacant place can be filled in 5 different ways by any one of the 5 flags following which, the lower vacant place can be filled in 4 different ways by any one Of the remaining 4 different flags.
Thus, by multiplication principle, the number of different signals that can be generated is 5 × 4 = 20.