11th NCERT Permutations And Combinations
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Miscellaneous Exercise 7 Questions 11

Question (1)

How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?

Solution

In the word Daughter, there are 3 vowels out of which 2 has to select it can be done in 3C2 different ways.
And out of 5 consonants 3 has to be selected , it can be done in 5C3 different ways.
Now 5 letters are selected has to be arranged in a row to form the word. It can be arranged in 5! ways.
So the number of words formed with or without meaning
= 3 C2 × 5 C3 × 5!
= 3 × 10 × 120
= 3600

Question (2)

How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Solution

In the word 'EQUATION' there are 8 letters in which 5 vowels and 3 consonants. As we required all vowels and all consonants together we make two groups.
These groups can be arranged in a row in 2! different ways.
In a group of vowels there are 5 vowels. they can be rearranged in 5! ways.
While in group of consonants there are 3 consonants, they can be rearranged in 3! ways.
Number of words formed with or without meaning using the letters of word 'EQUATION'
= 2! × 5! × 3!
= 2 × 120 × 6
= 1440

Question (3)

A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls? (ii) atleast 3 girls? (iii) atmost 3 girls?

Solution

A committee of 7 has to be formed from 9 boys and 4 girls.
(i) Exactly 3 girls.
In a committee of 7 , there are 3 girls and 4 boys.
We have to select 3 girls out of 4, it can be done in 4C3 different ways.
And 4 boys have to be selected from 9 boys, it can be done in 9C4 ways.
Number of committee formed
$= 4{C_3} \times 9{C_4}$ $= 4 \times \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$ $= 504$ (ii) Atleast 3 girls
That is there are 3 or more than 3 girls in a committee.
So committee consist of 3girls and 4 boys or 4 girls and 3 boys.
$= 4{C_3} \times 9{C_4} + 4{C_4} \times 9{C_3}$ $= 4 \times \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2}} + 1 \times \frac{{9 \times 8 \times 7}}{{3 \times 2}}$ $= 504 + 84$ $= 588$ (iii) atmost 3 girls
The committee should have maximum 3 girls.
= total number of ways the committee formed - no. of committee with 4 girls.
$= 13{C_7} - 4{C_4} \times 9{C_3}$ $= \frac{{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7}}{{7 \times 6 \times 5 \times 4 \times 3 \times 2}} - 1 \times \frac{{9 \times 8 \times 7}}{{3 \times 2}}$ $= 1716 - 84$ $= 1632$

Question (4)

If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?

Solution

The word ' EXAMINATION' consist of 11 letters in which 'A', 'I' and 'N' are repeated 2 times.
Total words formed using them once is
$= \frac{{11!}}{{2!2!2!}}$ We arrange the letters dictionary order as A,A,E,I, I, M, N. N. O, T, x.
The number of words in a list before the first word starting with E are the words starting with A.
Let the first word be A, now there are 10 positions left to form the word.
It can be filled by 10! ways. But 'I' and 'N' are repeated twice.
The number of word before E
$= \frac{{10!}}{{2!2!}}$ $= \frac{{10 \times 9 \times 8 \times 5040}}{{2 \times 2}}$ $= 907200$

Question (5)

How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?

Solution

Using the digits 0, 1, 3, 5, 7, and 9 we have to form 6 digit number divisible by 10.
As the number to be divisible by 10 its unit place must have digit '0'.
So there is only one possibility for unit place.
No digits are repeated , so for 5 places there are 5 digits to be arranged to form a number.
It can be done in 5! ways.
So number formed = 5!
= 120

Question (6)

The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

Solution

To form words with two different vowels and 2 different consonants, we will select 2 vowels from 5 vowels.
It can be done in 5C2 different ways.
Now we will select 2 different consonants from 21, it can be done in 21C2 different ways.
After selections of letters they have to be arranged in a row to form the word. it can be done in 4! ways.
The number of words formed
$= 5{C_2} \times 21{C_2} \times 4!$ $= \frac{{5 \times 4}}{2} \times \frac{{21 \times 20}}{2} \times 24$ $= 50400$

Question (7)

In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Solution

In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively.
A student is required to attempt 8 questions in all, selecting at least 3 from each part.
Possibilites are 3 from I and 5 from II OR 4 from I and 4 from II OR 5 from I and 3 from II
The number of ways the selection is done =
$= 5{C_3} \times 7{C_5} + 5{C_4} \times 7{C_4} + 5{C_5} \times 7{C_3}$ $= 10 \times 21 + 5 \times 35 + 1 \times 35$ $= 210 + 175 + 35$ $= 420$

Question (8)

Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Solution

In 5 cards there are exactly 1 kings.
There are 4 kings out of which 1 are to be selected in 4C1 ways.
Now out of 48 cards we have to select 4 cards. it can be done in 48C4 ways.
The number of combinations possible
$= 4{C_1} \times 48{C_4}$

Question (9)

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

Solution

There are 5 men and 4 women totally 9 persons have to be arranged in row , such that women occupy the even places.
There are 4 even positions in a row that has to be occupied by 4 women . it can be done in 4! ways.
Now remaining 5 positions will be occupied by 5 men, the arrangement will be done in 5! ways.
So the number of arrangements possible
$= 4! \times 5!$ $= 24 \times 120$ $= 2880$

Question (10)

From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

Solution

There are 2 possibilities (i) all 3 will go to the party (ii) none of these will go to party.
(i) If all 3 join the party , so now out of remaining 22 we have to select 7 .
It can be done in 22C7 ways.
(ii) If none of them join the party,
So out of 22 we have to select 10, it can be done in 22C10 ways.
So total ways to select =
$= 22{C_7} + 22{C_{10}}$

Question (11)

In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?

Solution

In the given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once.
Now we require that all 'S' are together.
So we will make a group of all 'S' , and count as one unit.
In a word there are 13 letters in that 4 'S' has considered as a one unit. so remaining 9 will be considered as 9 different units.
There are 10 units that has to be arranged in a row. It can be done by 10! ways.
In this 10 units 'A' is repeated 3 times, 'I' 2 times and 'N' 2 times.
So the number of arrangements of these letters in a row
$= \frac{{10!}}{{3!2!2!}}$ $= \frac{{10 \times 9 \times 8 \times 7 \times 6 \times 120}}{{6 \times 2 \times 2}}$ $= 151200$