11th NCERT Permutations And Combinations
Hi

## Exercise 7.3 Questions 11

Question (1)

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Solution

To form 3 digit number using the digits 1 to 9 without repeataion of the digits.
To fill 3 positions out of 9, so it can be done in 9P3 ways.
The numbers formed
= 9 P3
= 9 × 8 × 7
= 504

Question (2)

How many 4-digit numbers are there with no digit repeated?

Solution

We have to form 4 digit number without repeation using all digits from 0 to 9
we can not select 0 at thousand place By selecting 0 number will be 3 digit.
So we can select any one out of 9 digit . So we have 9 options for thousand place.
The digit selected at thousand place can not be repeated again. so now we have 9 digits left.
To form next three digit number we give the 3 positios from 9 digits, it can be done in 9P3 different ways.
So 4 digit number formed
= 9 × 9P3
= 9 × 9×8× 7
= 4536

Question (3)

How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Solution

To get even number digit at unit place must be even.
In given digits 3 digits are even.
So at unit place we will selct any one of these 3 digits, it can be done in 3 different ways.
Now the digit selected at unit place cannot be repeated,
So at ten and hundred place we have to select any two digits out of 5 digits.
the selection can be done in 5P2 different ways.
So 3 digits even numbers formed using the digits 1, 2, 3, 4, 6, 7 without repetation
= 3 × 5P2
= 3 × 5 × 4
= 60.

Question (4)

Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

Solution

To form 4 digit numbers using digits 1, 2, 3, 4, 5 without repetation.
we have to give the positions for digits, it can be done in 5P4 ways.
So number formed =
= 5P4
= 5 × 4× 3× 2
= 120
To form even number we have to select even digits at unit place.
There are 2 even digits. we will select any one at unit place, so we have 2 options for unit place.
Now out of remaining 4 we will fill up the three positions to form 4 digit number. It can be done by 4P3 ways.
So even numbers formed using the given digits with out repetation
= 2 × 4P3
= 2 × 4 × 3× 2
= 48.

Question (5)

From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

Solution

the two positions chairman and vice chairman have to be selected from 8 persons, it can be selected by 8P2 ways.
The number of ways to select these positions
= 8P2
= 8 × 7
= 56.

Question (6)

Find n if $n - 1{P_3}:n{P_4} = 1:9$

Solution

$n - 1{P_3}:n{P_4} = 1:9$ $\frac{{n - 1{P_3}}}{{n{P_4}}} = \frac{1}{9}$ $\frac{{\left( {n - 1} \right)!}}{{\left( {n - 4} \right)!}} \times \frac{{\left( {n - 4} \right)!}}{{n!}} = \frac{1}{9}$ $\frac{{\left( {n - 1} \right)!}}{{n\left( {n - 1} \right)!}} = \frac{1}{9}$ $n = 9$

Question (7)

Find r if (i)$5{P_r} = 2\;\left( {6{P_{r - 1}}} \right)$ (ii) $5{P_r} = 6{P_{r - 1}}$

Solution

$5{P_r} = 2\;\left( {6{P_{r - 1}}} \right)$ $\frac{{5!}}{{\left( {5 - r} \right)!}} = 2\left[ {\frac{{6!}}{{\left( {6 - \left( {r - 1} \right)} \right)!}}} \right]$ $\frac{{5!}}{{\left( {5 - r} \right)!}} = 2 \times \frac{{6 \times 5!}}{{\left( {7 - r} \right)\left( {6 - r} \right)\left( {5 - r} \right)!}}$ $\left( {7 - r} \right)\left( {6 - r} \right) = 12$ $\left( {7 - r} \right)\left( {6 - r} \right) = 4 \times 3$ $7 - r = 4$ $r = 3$ (ii) $5{P_r} = 6{P_{r - 1}}$ $\frac{{5!}}{{\left( {5 - r} \right)!}} = \frac{{6!}}{{\left[ {6 - \left( {r - 1} \right)} \right]!}}$ $\frac{{5!}}{{\left( {5 - r} \right)!}} = \frac{{6 \times 5!}}{{\left( {7 - r} \right)\left( {6 - r} \right)\left( {5 - r} \right)!}}$ $\left( {7 - r} \right)\left( {6 - r} \right) = 6$ $\left( {7 - r} \right)\left( {6 - r} \right) = 3 \times 2$ $7 - r = 3$ $r = 4$

Question (8)

How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Solution

The word 'EQUATION' consist of 8 letters.
We have to use all letters once to form the words.
We have to arrange these 8 letters in a row.
It can be done in 8! ways.
So number of word formed with or without meaning using all letters once
= 8!
= 8 ×7 ×6 × 5 × 4 × 3 × 2
= 40320

Question (9)

How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time , ( ii) all letters are used at a time, (iii)All letters are used but first letter is a vowel?

Solution

The word 'MONDAY' consist of 6 letters.
(i) 4 letters are used at a time to form the word
As we have to use 4 letters at a time it means 4 letter word has to formed.
To fill 4 positions there are 6 letters , it can be done in 6P4 ways.
Words formed = 6P4
= 6 × 5 × 4 × 3
= 360
(ii) All the letters are used at time.
So to form 6 letter word using 6 letters is arranging 6 letters in a row.
It can be done in 6! ways.
The number of words formed = 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720.
(iii) All letters are used but first letter is vowel.
In a word there are 2 vowels, so for first position there are 2 possibilities.
Now remaning 5 letters will be arranged in next 5 positions in 5! different ways.
Number of word formed = 2 × 5!
= 2 × 5 × 4 × 3 × 2 × 1
= 240.

Question (10)

In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

Solution

The word ' MISSISSIPPI' consist of 11 letters in which 'I' occures 4 times, 'S' occures 4 times, 'P' occures 2 times .
The total words formed using all letters once
$= \frac{{11!}}{{4!4!2!}}$ $= \frac{{11 \times 10 \times 9 \times 8 \times 7 \times 720}}{{24 \times 24 \times 2}}$ $= 34650$ Now we will form the word in which all 'I' are together.
So make a group of all 'I' and other induvisials are consider as different groups, there are 8 groups has to arranged in a row.
It has 'S' 4times and 'P' 2 times.
So number of word formed using all 'I' together
$= \frac{{8!}}{{4!2!}}$ $= \frac{{8 \times 7 \times 720}}{{24 \times 2}}$ $= 840$ The number of words formed with all 'I' not together
= total word formed - the number of words formed with all 'I' together.
= 34650 - 840
= 33810

Question (11)

In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (ii) there are always 4 letters between P and S?

Solution

The word ' PERMUTATIONS' consist of 12 letters, in which 'T' occures 2 times,
(i) Word is start with P and ends with S. so these positions are fixed.
Now remaning letters are arranged in the 10 positions., in 10! ways but 'T' occues 2 times
So the word formed with starting with P and ending with S
= $= 1 \times 1 \times \frac{{10!}}{{2!}}$ $= \frac{{10 \times 9 \times 8 \times 7 \times 720}}{2}$ $= 1814400$ (ii) Vowels are all together.
There are 5 vowels in the word. As all vowels are required together, Make a group of them.
Remaning consonants will be considered as different induvisials. so there are 8 groups that has to be arranged in a row.
It can be done in 8! ways. but 'T' occures 2 times so we have to divide by 2!.
In a group of vowels , 5 vowels can rearranged in 5! ways.
So the number of words formed
$= 5! \times \frac{{8!}}{{2!}}$ $= 120 \times \frac{{8 \times 7 \times 720}}{2}$ $= 2419200$ (iii) There are always 4 letters between P and S.
Out of 10 letters first we will select 4 letters. it can be done 10C4 different ways.
Then we will arrange them in a row it can be done in 4! ways.
Now P and S can also re arrange themselves in 2 ways.
Now there is a group of 6 letters including P and S. other letters are considered as induvidual groups,
Now there are now 7 groups has to be arranged in a row it can be done in 7! ways.
But 'T' occures 2 times, so we have to divide by 2.
The number of word formed by 4 letters between P and S
$= 10{C_4} \times 4! \times 2 \times \frac{{7!}}{{2!}}$ $= \frac{{10 \times 9 \times 8 \times 7}}{{4 \times 3 \times 2}} \times 24 \times 2 \times \frac{{5040}}{2}$ $= 25401600$