11th NCERT Permutations And Combinations
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## Exercise 7.1 Questions 6

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?

Solution

Using the digits 1, 2, 3, 4 and 5 we have to form 3 digit number,
(i) Repetation is allowed.
First we will select the digit at hundred place out of 5 given digits. we have 5 options for it.
As the repeatation is allowed we have 5 options for tenth place and for unit place as well.
So the number formed =
= 5 × 5 × 5
= 125.
(ii) repetation is not allowed.
First we will select the digit at hundred place out of 5 given digits. we have 5 options for it.
As the repetation is not allowed, the digit selected at hundredth place will not be repeated.
So we have to select the digit out of 4 options, so we have 4 options.
The digit selected at tenth place will also not be repeated.So for the unit place we have to select the digit out of 3.
The numbers formed
= 5 × 4 × 3
= 60.

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?

Solution

To form the three digit even number using the digits 1, 2, 3, 4, 5, and 6 , the digit at unit place must be even.
We have to select one digit out of 2, 4, and 6. It can be done in 3 different ways.
Now repetation of the digits are allowed. So we have 6 possibilites for tenth and hundred place.
So the even number formed with repetation
= 6 × 6 × 3
= 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?

Solution

To form 4 letter code we have to use first 10 letters of English and no letter will be repeated.
So give 4 positions from 10, it can be done 10P4 different ways.
The number of codes form
= 10P4
= 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?

Solution

To form 5–digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67.
As the digit can not be repeated , digits 6 and 7 can not be used again in next three digit.
So we have to arrange 3 digits out of remaning 8 digits to form a number , if can be done in 8P3 different ways.
So the number formed
= 8P3
= 8 × 7 × 6
= 336

Question (5)

A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Solution

A coin has to options , either head or tail appears.
When 3 coins are tossed for every coin it has 2 options.
So number of possible outcomes
= 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?

Solution

There are 5 flag of different colours are given . Using to flag one below the other we have to form single.
It can be done in 5P2 different ways.
The number of singles formed
= 5P2
= 5 × 4
= 20