GSEB Class 11 Maths Solutions Chapter 7 Permutations and Combinations Exercise 7.1

Question 1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5, assuming.
(i) repetition of digits is allowed?
(ii) repetition of digits is not allowed?
Answer:
(i) There are five numbers available: 1, 2, 3, 4, and 5. Each number can be chosen any number of times. This means we can pick the first number in 5 different ways. The second number can also be picked in 5 ways, and the third number can be picked in 5 ways. Thus, the total count of ways to choose three numbers is \( 5 \times 5 \times 5 = 125 \) ways.
(ii) With the condition that numbers cannot repeat, the first number can be chosen in 5 ways. After picking the first number, four numbers remain. The second number can then be chosen in 4 ways. After that, three numbers are left, so the third number can be chosen in 3 ways. Therefore, the total count of ways to choose three numbers is \( 5 \times 4 \times 3 = 60 \).
In simple words: When digits can repeat, you have 5 choices for each of the three places, so \( 5 \times 5 \times 5 = 125 \) numbers. If digits cannot repeat, you have 5 choices for the first place, 4 for the second, and 3 for the third, making \( 5 \times 4 \times 3 = 60 \) numbers.

Exam Tip: For problems involving combinations and permutations, clearly distinguish between situations where repetition is allowed and where it is not, as this directly affects the number of choices for each position.

Question 2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6, if the digits can be repeated?
Answer: To create a 3-digit even number using digits 1, 2, 3, 4, 5, 6 with repetition allowed, the unit's place must be an even digit (2, 4, or 6). If 2 is fixed at the unit's place, the ten's place can be filled in 6 ways (any of the 6 digits can be used). The hundred's place can also be filled in 6 ways. So, 36 numbers can be formed when 2 is in the unit's place (\( 6 \times 6 = 36 \)). Similarly, 36 numbers can be formed when 4 is in the unit's place, and 36 numbers when 6 is in the unit's place. Therefore, the total number of 3-digit even numbers possible with repetition is \( 36 \times 3 = 108 \).
In simple words: To make an even number, the last digit must be even. Since repetition is allowed, for each even digit (2, 4, 6) in the last place, there are 6 choices for the middle digit and 6 choices for the first digit. Multiply \( 6 \times 6 \) for each even option, then add those results together.

Exam Tip: When forming numbers with specific properties (like being even or odd), always start by considering the restricted position (e.g., the unit's place for even/odd numbers) first.

Question 3. How many 4-letter code words are possible, using the first 10 letters of the English alphabet, if no letter can be repeated?
Answer: To form a 4-letter code word from the first 10 English letters without repeating any letter, we consider the choices for each position. The first letter of the code word can be chosen in 10 different ways. After picking the first letter, 9 letters are left. So, the second letter can be chosen in 9 ways. Following this, the third letter can be chosen in 8 ways, and the fourth letter can be chosen in 7 ways. Using the Fundamental Principle of Counting, the total number of ways to choose four letters from 10 distinct letters is \( 10 \times 9 \times 8 \times 7 = 5040 \) ways.
In simple words: For a 4-letter code with no repeats from 10 letters, you pick the first letter in 10 ways, the second in 9 ways, the third in 8 ways, and the fourth in 7 ways. Multiply these choices together to get the total number of codes.

Exam Tip: When letters or digits cannot be repeated, the number of choices decreases by one for each subsequent position.

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: There are 10 total digits, from 0 to 9. The 5-digit telephone number must start with '67'. Since no digit can appear more than once, the digits 6 and 7 are already used for the first two positions. This means 2 digits are gone from the initial 10 digits. We have \( 10 - 2 = 8 \) digits remaining. We need to fill the next three places (the 3rd, 4th, and 5th digits).
The third place (after 67) can be filled in 8 ways (from the remaining 8 digits).
The fourth place can be filled in 7 ways (since one more digit has been used).
The fifth place can be filled in 6 ways (since another digit has been used).
Therefore, the total number of 5-digit telephone numbers that can be created is \( 8 \times 7 \times 6 = 336 \).
In simple words: Since the first two digits are fixed as 67 and no digits can repeat, we have 8 digits left to choose from for the next three spots. The third digit has 8 options, the fourth has 7, and the fifth has 6. Multiply these numbers to find the total possible phone numbers.

Exam Tip: For fixed starting digits, subtract them from the total available digits before calculating the remaining choices. This ensures no repetition if that condition is stated.

Question 5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Answer: When you toss a coin, there are two potential outcomes: Head (H) or Tail (T). Similarly, for the second toss, there are again two possible outcomes. The same two outcomes will also appear for the third toss. According to the Fundamental Principle of Counting (FPC), the total number of outcomes when tossing a coin three times is found by multiplying the number of outcomes for each toss. Thus, the total number of outcomes is \( 2 \times 2 \times 2 \) ways, which equals 8 ways.
In simple words: Each coin flip has 2 possible results (Heads or Tails). If you flip it 3 times, you multiply the possibilities for each flip: \( 2 \times 2 \times 2 \), which gives you 8 total outcomes.

Exam Tip: For independent events like coin tosses, the total number of outcomes is the product of the number of outcomes for each individual event.

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: To generate signals using 2 flags, one below the other, from a set of 5 different coloured flags, we consider the choices for each position. The first position (top flag) can be filled in 5 different ways, using any of the 5 flags. After placing the first flag, 4 flags are left. So, the second position (bottom flag) can be filled in 4 ways. Therefore, the total number of unique signals that can be generated is \( 5 \times 4 = 20 \).
In simple words: You have 5 flags for the top position. Once one is chosen, you have 4 flags left for the bottom position. Multiplying these choices gives the total number of distinct signals: \( 5 \times 4 = 20 \).

Exam Tip: This is a permutation problem without repetition, as choosing a flag for the top position means it cannot be chosen for the bottom position.