Permutation and Combination – Exercise 4.5 – Class X

Previous Exercise – Permutation and Combination – Exercise 4.4


Permutation and Combination – Exercise 4.5

  1. How many words with or without dictionary meaning can be formed using all the letters of the word ‘JOULE’ using each letters exactly once?
  2. It is required to seat 5 men and 4 women in a row so that the women occupy even places. How many such arrangements are possible?
  3. In how many ways can 6 women draw water from 6 wells, if no well remains unused?
  4. 8 students are participating in a competition. In how many ways can the forst three prizes be won?
  5. Find the total number of 2 – digits numbers.
  6. How many 4 – digit numbers can be formed using the digits 1, 2, 3, 7, 8, 9(repeating not allowed)?

(a) How many of these are less than 6000?

(b)How many of these are even?

  1. There are 15 buses running between two towns. In how many ways can a man go to one town and return by different bus?

Permutation and Combination – Exercise 4.5 – Solutions

  1. How many words with or without dictionary meaning can be formed using all the letters of the word ‘JOULE’ using each letters exactly once?

Solution:

Here n = 5 and r = 5

Then, we have nPr = 5P5 = 5!/(5-5)! = 5!/0! = 5!/1 = 5! = 120.

Therefore, 120 with or without dictionary meaning can be formed using all the letters of the word ‘JOULE’ using each letters exactly once


  1. It is required to seat 5 men and 4 women in a row so that the women occupy even places. How many such arrangements are possible?

Solution:

It is required to seat 5 men and 4 women in a row so that the women occupy even places.

Then, 5! x 4! = (1 x 2 x 3 x 4 x 5) x (1 x 2 x 3 x 4) = 120 x 24 = 2880

Hence, 2880 such arrangements are possible.


  1. In how many ways can 6 women draw water from 6 wells, if no well remains unused?

Solution:

Here n = 6 , r = 6

nPr = 6P6 = 6!/(6-6)! = 6!/0! = 6!/1 =  6! = 720

In 720 ways can 6 women draw water from 6 wells, if no well remains unused.


  1. 8 students are participating in a competition. In how many ways can first three prizes be won?

Solution:

Here, n = 8 and r = 3

nPr = 8P3 = 8!/(8-3)! = 8!/5! = 5! x 6 x 7 x 8/5! = 6 x 7 x 8 = 336

Therefore, in 336 ways can first three prizes be won.


  1. Find the total number of 2 – digits numbers.

Solution:

We have 99 number from 1 to 99, but from digits from 1 to 9 are single digits. Therefore we have 99 – 9 = 90 .

Hence, in total 90 two digit numbers.


  1. How many 4 – digit numbers can be formed using the digits 1, 2, 3, 7, 8, 9(repeating not allowed)?

(a) How many of these are less than 6000?

(b)How many of these are even?

( c) How many of these end with 7?

Solution:

(a)In order to be less than 4000 you can only use 1, 2, 3 as the first digit so as to get the number less than 6000.

This leaves us with five digits to choose from for the second digit(as 6 digits are given and repetition is not allowed). In ten’s place we have four digits to chose(in order to avoid repetition) and at unit’s place we have 3 digits left to chose from.

Thousand’s placeHundred’s placeTen’s placeUnit’s place
3 ways(1, 2, 3)5 ways4 ways3 ways

So overall we have 3 x 5 x 4 x 3 = 180 numbers.

(b)To form an even number it should end with even numbers. Here, number should end with 2 and 8.This leaves us with five digits to choose from for the thousand’s place(as 6 digits are given and repetition is not allowed). In hundred’s place we have four digits to chose(in order to avoid repetition) and at ten’s place we have 3 digits left to chose from.

Thousand’s placeHundred’s placeTen’s placeUnit’s place
5 ways4 ways3 ways2 ways

So overall we have 5 x 4 x 3 x 2 = 120 numbers.

(c) It is given that the number should end with 7. This leaves us with five digits to choose from for the thousand’s place(as 6 digits are given and repetition is not allowed). In hundred’s place we have four digits to chose(in order to avoid repetition) and at ten’s place we have 3 digits left to chose from.

Thousand’s placeHundred’s placeTen’s placeUnit’s place
5 ways4 ways3 ways2 ways

So overall we have 5 x 4 x 3 x 1 = 60 numbers.


  1. There are 15 buses running between two towns. In how many ways can a man go to one town and return by different bus?

Solution:

Given, there are 15 buses running between two towns.

n = 15 and r = 2

Then nPr = 15P2 = 15!/(15-2)! = 13! x 14 x 15/13! = 14 x 15 = 210

Therefore, there are 210 ways a man can go to one town and return by different bus.


Next exercise – Permutation and Combination – Exercise 4.5


 

Leave a Reply

Your email address will not be published. Required fields are marked *