Chapter 7: Permutations and Combinations Miscellaneous Exercise

The word DAUGHTER has 3 vowels, A, E, U and 5 consonants D, G, H, T and R.

The three vowels can be chosen in ³C₂, as only two vowels are to be chosen.

Similarly, the five consonants can be chosen in ⁵C₃ ways.

∴ The number of choosing 2 vowels and 5 consonants would be ³C₂ ×⁵C₃

= 30

∴ The total number of ways is 30.

Each of these 5 letters can be arranged in 5 ways to form different words = ⁵P₅

The total number of words formed would be = 30 × 120 = 3600

In the word EQUATION, there are 5 vowels (A, E, I, O, U) and 3 consonants (Q, T, N)

The numbers of ways in which 5 vowels can be arranged are ⁵C₅

…………… (i)

Similarly, the numbers of ways in which 3 consonants can be arranged are ³P₃

…………….. (ii)

There are two ways in which vowels and consonants can appear together:

(AEIOU) (QTN) or (QTN) (AEIOU)

∴ The total number of ways in which vowel and consonant can appear together are 2 × ⁵C₅ × ³C₃

∴ 2 × 120 × 6 = 1440

(i) Given, exactly 3 girls.

The total numbers of girls are 4.

Out of which, 3 are to be chosen.

∴ The number of ways in which choice would be made = ⁴C₃

The number of boys is 9, out of which 4 are to be chosen, which is given by ⁹C₄

Total ways of forming the committee with exactly three girls

= ⁴C₃ × ⁹C₄

= 

(ii) Given, at least 3 girls.

There are two possibilities for making a committee choosing at least 3 girls.

There are 3 girls and 4 boys, or there are 4 girls and 3 boys.

Choosing three girls we have done in (I).

Choosing four girls and 3 boys would be done in ⁴C₄ ways.

And choosing 3 boys would be done in ⁹C₃

Total ways = ⁴C₄ ×⁹C₃

The total number of ways of making the committee are

504 + 84 = 588

(iii) Given, at most 3 girls.

In this case, the numbers of possibilities are

0 girls and 7 boys

1 girl and 6 boys

2 girls and 5 boys

3 girls and 4 boys

The number of ways to choose 0 girls and 7 boys = ⁴C₀ × ⁹C₇

The number of choosing 3 girls and 4 boys is done in (1)

= 504

The total number of ways in which a committee can have at most 3 girls are = 36 + 336 + 756 + 504 = 1632

In a dictionary, words are listed alphabetically, so to find the words,

Listed before E should start with the letter either A, B, C or D.

But the word EXAMINATION doesn’t have B, C or D.

Hence, the words should start with the letter A.

The remaining 10 places are to be filled by the remaining letters of the word EXAMINATION, which are E, X, A, M, 2N, T, 2I, 0.

Since the letters are repeating, the formula used would be

Where n is the remaining number of letters, p₁ and p₂ are the number of times the repeated terms occur.

The number of words in the list before the word starting with E

= words starting with letter A = 907200

The number is divisible by 10 if the unit place has 0 in it.

The 6-digit number is to be formed out of which unit place is fixed as 0.

The remaining 5 places can be filled by 1, 3, 5, 7 and 9.

Here, n = 5

And the numbers of choice available are 5.

So, the total ways in which the rest of the places can be filled are ⁵P₅

We know that there are 5 vowels and 21 consonants in the English alphabet.

Choosing two vowels out of 5 would be done in ⁵C₂ ways.

Choosing 2 consonants out of 21 can be done in ²¹C₂ ways.

The total number of ways to select 2 vowels and 2 consonants

= ⁵C₂ × ²¹C₂

Each of these four letters can be arranged in four ways ⁴P₄

The total numbers of words that can be formed are

24 × 2100 = 50400

The student can choose 3 questions from part I and 5 from part II,

Or

4 questions from part I and 4 from part II,

5 questions from parts 1 and 3 from part II.

We have a deck of cards that has 4 kings.

The numbers of remaining cards are 52.

Ways of selecting a king from the deck = ⁴C₁

Ways of selecting the remaining 4 cards from 48 cards= ⁴⁸C₄

The total number of selecting the 5 cards having one king always

= ⁴C₁ × ⁴⁸C₄

Given, there is a total of 9 people.

Women occupy even places, which means they will be sitting in 2^nd, 4^th, 6^th and 8^th places, whereas men will be sitting in 1^st, 3^rd, 5^th,7^th and 9^th places.

4 women can sit in four places and ways they can be seated= ⁴P₄

5 men can occupy 5 seats in 5 ways.

The number of ways in which these can be seated = ⁵P₅

The total number of sitting arrangements possible =

24 × 120 = 2880

In this question, we get 2 options that are

(i) Either all 3 will go

The remaining students in the class are: 25 – 3 = 22

The number of students that remain to be chosen for the party = 7

The number of ways to choose the remaining 22 students = ²²C₇

=

(ii) None of them will go.

The students going will be 10.

Remaining students eligible for going = 22

The number of ways in which these 10 students can be selected is ²²C₁₀

The total number of ways in which students can be chosen = 170544 + 646646 = 817190