### Question 1. In how many ways can the letters of the word ‘FAILURE’ be arranged so that the consonants may occupy only odd positions?

**Solution:**

From the given word, we get

F, L, R = 3 consonants

A, I, U, E = 4 vowels

1, 3, 5, 7 = 4 odd positions possible

Out of these 4 positions select 3 positions for the 3 given consonants =

^{4}P_{3}= 4 waysNow for these positions, ways to arrange consonants = 3! = 6

And ways to arrange the vowels in remaining positions are = 4! = 24

Total ways = 4 x 6 x 24 = 576

### Question 2. In how many ways can the letters of the word ‘STRANGE’ be arranged so that

### (i) The vowels come together?

### (ii) The vowels never come together? And

### (iii) The vowels occupy only the odd places?

**Solution:**

(i)2 vowels are given, for these to be together let the first of two vowels occupy a position pNow, there are 6 ways in which p can be chosen.

(All 7 positions except the last position, because if the first vowel

comes at the last position, no position will be left there for the second vowel)

For each p chosen, the vowels can be put in two orders (AE, EA) = 2 ways

Remaining letters can be arranged in = 5! = 120 ways

Total ways = Ways of choosing positions for vowels x

Ways to arrange the vowels x

Ways to arrange the remaining letters

= 6 x 2 x 120 = 1440

(ii)Let the first vowel be at position p.p can be chosen in 6 different ways (1 to 6).

When first vowel is at p, then second can be from p+2 to 7 so that not consecutive.

= 7 – (p + 2) + 1 = (6 – p)ways

Case 1 : p = 1 ⇒ 6 – p = 5

Case 2: p=2 ⇒ 6 – p = 4

and so on till 6 – p = 0 for Case 6.

No. of ways of selecting vowels positions in the word = 5 + 4 + 3 + 2 + 1 + 0 = 15

For each selection, ways to put the vowels = 2 (AE and EA)

Ways of putting the remaining letters = 5! = 120

Total ways = 15 x 2 x 120 = 3600

(iii)Choosing 2 odd positions from 4 =^{4}P_{2 }= 4 x 3 / 2 = 6Ways of arranging two vowels in selected positions = 2! = 2

Ways of arranging other letters are = 5! = 120

Total ways = 6 x 2 x 120 = 1440

### Question 3. How many words can be formed from the letters of the word ‘SUNDAY’? How many of these begin with D?

**Solution:**

Total number of letters present in the given word = 6

So, ways to arrange 6 letters to get the words = 6! = 720

Now, when D is fixed at position 1, remaining 5 can be arranged in ways = 5! = 120

### Question 4. How many words can be formed out of the letters of the word, ‘ORIENTAL,’ so that the vowels always occupy the odd places?

**Solution:**

From the given word, we get 4 vowels and 4 odd positions possible (1, 3, 5, 7)

So, the ways to arrange the vowels = 4! = 24

Ways to arrange consonant letters = 4! = 24

Total ways = 24 x 24 = 576

### Question 5. How many different words can be formed with the letters of word ‘SUNDAY’? How many of the words begin with N? How many begin with N and end in Y?

**Solution:**

Total number of letters present in the given word = 6

Now, Ways to arrange the 6 letters to get the words = 6! = 720

So now we find how many words begin with N

Let the fix N on the first position and remaining letter is 5

So the arrangement of 5 letters can be in 5! ways = 120 ways

Now find how many begin with N and end in Y

Let fix Y at position number 6 and N at position number 1 and the remaining letter is 4

So, the arrangement of 4 letters can be in 4! ways = 24 ways

### Question 6. How many different words can be formed from the letters of the word ‘GANESHPURI’? In how many of these words:

### (i) The letter G always occupies the first place?

### (ii) The letter P and I respectively occupy the first and last place?

### (iii) Are the vowels always together?

### (iv) The vowels always occupy even places?

**Solution:**

Given:Total number of letters present in the given word = 10

(i)Fix G at position number 1 and the remaining letters are 9So, the remaining letters can be arranged 9! = 51840 ways

(ii)The letter P and I respectively occupy the first and last place and the remaining letters are 8So, the remaining letters can be arranged 8! = 5760 ways

(iii)Total number of vowels present = 4The first vowel can be at max at position 7 (otherwise remaining vowels would

not have enough required positions)

So from 1 to 7 positions first vowel can be put at

^{7}P_{1 }= 7 waysRemaining will be put in next consecutive positions in the word. (so 1 way)

Ways to select the vowels positions = Ways to put first vowel x Ways to put remaining vowels

= 7 x 1 = 7 ways

Ways to arrange the 4 vowels in these 4 selected positions = 4! = 24 ways

Ways to arrange the consonant letters will be = 6! = 720 ways

Total words having vowels always together = Ways to choose the vowels positions x

Ways to arrange vowels x

Ways to arrange consonants

= 7 x 24 x 720

= 120960

(iv)5 positions possible that are even (2, 4, 6, 8, 10)We have 4 vowels and need to choose 4 even positions =

^{5}P_{4}= 5 waysWays to arrange these 4 vowels = 4! = 24

Ways to arrange the 6 consonants = 6! = 720

Total words = 5 x 24 x 720

= 86400

### Question 7. How many permutations can be formed by the letters of the word, ‘VOWELS’, when

### (i) There is no restriction on letters?

### (ii) Each word begins with E?

### (iii) Each word begins with O and ends with L?

### (iv) All vowels come together?

### (v) All consonants come together?

**Solution:**

Given:Total number of letters = 6

(i)Ways to arrange 6 letters are = 6! = 720

(ii)Fix E at first position and the remaining letters are 5So, the remaining letters can be arranged = 5! = 120 ways

(iii)Fix O at 1st and L at last position. So the remaining 4 lettersThese 4 letters can be arranged = 4! = 24 ways

(iv)2 vowels together can consider them 1 symbol = 2 ways for this new symbol (EO and OE)1 symbol and 4 letters = 5 things are to be arranged so number of ways = 5! =120

Total ways = 120 x 2 = 240

(v)Consonants together = VWLS are 1 group = 4! waysGroup of consonants and 2 vowels = 3 things = 3! ways to arrange

Total ways = 4! x 3! = 144

### Question 8. How many words can be formed out of the letters of the word ‘ARTICLE’, so that vowels occupy even places?

**Solution:**

In the given word, we have 3 vowels and also 3 even positions there

So 3! ways to occupy places for vowels.

For consonants = 4! = 24 ways

Total words = 3! x 4! = 144

### Question 9. In how many ways can a lawn tennis mixed double be made up from seven married couples if no husband and wife play in the same set?

**Solution:**

2 teams with 1 man and 1 woman

Need to select 2 man so number of ways are =

^{7}P_{2}= 21Need to select 2 woman out of 5 (except wives of selected men), ways =

^{5}P_{2}= 10Now, ways to assign teams to men are = 2 (1st man to team A or to team B)

Similarly, for women too = 2 ways

Number of required ways = 21 x 10 x 2 x 2 = 840

### Question 10. m men and n women are to be seated in a row so that no two women sit together. If m>n then show that the number of ways in which they can be seated as

**Solution:**

Let us first seat the men = m! ways

Now, m+1 spaces are created for women tor sit.

(before 1st man, between two consecutive men, after last seated man)

Out of these m+1 spaces, we need to choose n spaces for the women =

^{m+1}P_{n}possible waysArrange women in n! ways after selecting seats for them above.

Total ways to seat = m! x n! x (m+1)! / ( (m + 1 – n!) x n! )

= m! (m+1)! / (m+1-n)!

Hence, shown.

### Question 11. How many words (with or without dictionary meaning) can be made from the letters in 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 is vowel?

**Solution:**

(i)Select 4 letters out of 6 =^{6}P_{4}= 6 x 5 / 2 = 15Ways to arrange these = 4! = 24

Total such words possible = 15 x 24 = 360

(ii)Number of such words = 6! = 720

(iii)Possibilities for first position = 2 (O or A)Number of arrangements for other positions in the word are = 5!

Total words = 2 x 5! = 240

### Question 12. How many three-letter words can be made using the letters of the word ‘ORIENTAL’?

**Solution:**

Total number of the letters present in the given word = 4

Select 3 letters out of 8 =

^{8}P_{3 }= 56 waysWe have 3! ways to arrange

Hence, the total words = 56 x 3! = 336