# Permutation and Combination – Exercise 4.1 – Class X

#### Permutation and Combination – Exercise 4.1

1. How many 3 –digit numbers can be formed using the digits 1, 2, 3, 4, 4, 5, 6 without repeating any digit?
2. How many 3 – digit even numbers can be formed using the digits 3, 5, 7, 8, 9, if the digits are not repeated?
3. How many 3 letter code can be formed using the first 10 letters of English alphabet if no letter can be repeated?
4. How many 5 digit telephone numbers can be formed using the digits ) to 9, if the numbers starts with 65 and no digit appear more than once?
5. IF a coin is tossed 3 times, find the number of outcomes?
6. Given 5 flags of different colours, how many different signals can be generated each signal requires the use of 2 flags one below the other?

#### Permutation and Combination – Exercise 4.1 – Solutions:

1. How many 3 –digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 without repeating any digit?

Solution:

The digits in the selection set are {1, 2, 3, 4, 5, 6}

We have to form 3 digit numbers from given 6 digits without repeating, with all digits distinct from each other, are the permutations of the 6 digits taken 3 at a time, and therefore, there are 6*5*4 = 10 of them.

1. How many 3 – digit even numbers can be formed using the digits 3, 5, 7, 8, 9, if the digits are not repeated?

Solution:

The digits in the selection set are {3, 5, 7, 8, 9}

To be even, the number must end with even numbers only such as 0, 2, 4, 6, 8. We can construct all of even three digit number ending with 8(as its only even number in the given set) are the permutation of the remaining 4 digits taken 3 at a time.  So number of 3 – digit even number can be formed without repetition = 1*4*3 = 12.

1. How many 3 letter code can be formed using the first 10 letters of English alphabet if no letter can be repeated?

Solution:

. The digits in the selection set are the first 10 letters of English alphabet.

We have to form 3 letter code from given 10 letters without repeating, with all letters distinct from each other, are the permutations of the 10 digits taken 3 at a time, and therefore, there are 10*9*8 = 720 of them

1. How many 5 digit telephone numbers can be formed using the digits 0 to 9, if the numbers starts with 65 and no digit appear more than once?

Solution:

We have to form 5 digit telephone number from the digits 0 to 9 without repeating and the number must start from 65.

Given 10 digits in the selection set, but the number must start from 65. So we have to arrange remaining 8 digits without repeating. Thus, first letter can be chosen in 8 ways, second letter can be chosen only in 7 ways and third letter can be chosen in 6 ways. So the total number of combinations = 8*7*6 = 336

1. If a coin is tossed 3 times, find the number of outcomes?

Solution:

A coin has 2 faces. If it’s tossed three times then the number outcomes = 2 x 2 x 2 = 8

1. Given 5 flags of different colours, how many different signals can be generated each signal requires the use of 2 flags one below the other?

Solution:

The upper portion can be fitted with 5 colours and the lower portion can be fitted with 4 colours. Therefore, the total number of signals that can be generated using two flags one below the other = 5 x 4 = 20 signals.

Next exercise Permutation and Combination – Exercise 4.2