**Chapter 3 Permutations and Combination Ex 3.3**

## Chapter 3 Permutations and Combination Ex 3.3

**Question 1.**

∴ (n – 3) (n – 4) (n – 5) = 120

∴ (n – 3) (n – 4) (n – 5) = 6 × 5 × 4

Comparing on both sides, we get

n – 3 = 6

∴ n = 9

**Question 2.**

(m + n) (m + n – 1) = 56

Let m + n = t

t(t – 1) = 56

– t – 56 = 0

(t – 8) (t + 7) = 0

t = 8 or t = -7

m + n = 8 or m + n = -7

But m + n ≠ -7

∴ m + n = 8 ……(i)

(m – n) (m – n – 1) = 12

Let m – n = a

a(a – 1) = 12

– a – 12 = 0

(a – 4)(a + 3) = 0

a = 4 or a = -3

m – n = 4 or m – n = -3

But m – n ≠ -3

∴ m – n = 4 ……(ii)

Adding (i) and (ii), we get

2m = 12

∴ m = 6

Substituting m = 6 in (ii), we get

6 – n = 4

∴ n = 2

Solution:

(14 – r)(13 – r) = 8 × 7

Comparing on both sides, we get

14 – r = 8

∴ r = 6

Solution:

**Question 5.How many 4 letter words can be formed using letters in the word MADHURI, if (a) letters can be repeated (b) letters cannot be repeated.Solution:**

There are 7 letters in the word MADHURI.

(a) A 4 letter word is to be formed from the letters of the word MADHURI and repetition of letters is allowed.

∴ 1st letter can be filled in 7 ways.

2nd letter can be filled in 7 ways.

3rd letter can be filled in 7 ways.

4th letter can be filled in 7 ways.

∴ Total no. of ways a 4-letter word can be formed = 7 × 7 × 7 × 7 = 2401

**Question 6.Determine the number of arrangements of letters of the word ALGORITHM if(a) vowels are always together.(b) no two vowels are together.(c) consonants are at even positions.(d) O is the first and T is the last letter.Solution:**
There are 9 letters in the word ALGORITHM.

(a) When vowels are always together.

There are 3 vowels in the word ALGORITHM (i.e., A, I, O).

Let us consider these 3 vowels as one unit.

This unit with 6 other letters is to be arranged.

∴ The number of arrangement =

^{7}P

_{7}= 7! = 5040

3 vowels can be arranged among themselves in

^{3}P

_{3}= 3! = 6 ways.

∴ Required number of arrangements = 7! × 3!

= 5040 × 6

= 30240

(b) When no two vowels are together.

There are 6 consonants in the word ALGORITHM,

they can be arranged among themselves in ^{6}P_{6} = 6! = 720 ways.

Let consonants be denoted by C.

_C _C_ C _C_C_C

There are 7 places marked by ‘_’ in which 3 vowels can be arranged.

∴ Required number of arrangements = 720 × 210 = 151200

(c) When consonants are at even positions.

There are 4 even places and 6 consonants in the word ALGORITHM.

∴ Required number of arrangements = 360 × 120 = 43200

(d) When O is the first and T is the last letter.

All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T.

∴ Position of O and T are fixed.

∴ Other 7 letters can be arranged between O and T among themselves in ^{7}P_{7} = 7! = 5040 ways.

∴ Required number of arrangements = 5040

**Question 7.In a group photograph, 6 teachers and principal are in the first row and 18 students are in the second row. There are 12 boys and 6 girls among the students. If the middle position is reserved for the principal and if no two girls are together, find the number of arrangements.Solution:**

In 1st row, 6 teachers can be arranged among themselves in

^{6}P

_{6}= 6! ways.

In the 2nd row, 12 boys can be arranged among themselves in

^{12}P

_{12}= 12! ways.

No two girls are together.

So, there are 13 places formed by 12 boys in which 6 girls occupy any 6 places in

^{13}P

_{6}ways.

∴ Required number of arrangements

**Question 8.Find the number of ways so that letters of the word HISTORY can be arranged as(a) Y and T are together(b) Y is next to T(c) there is no restriction(d) begin and end with a vowel(e) end in ST(f) begin with S and end with TSolution:**

There are 7 letters in the word HISTORY

(a) When ‘Y’ and ‘T’ are together.

Let us consider ‘Y’ and ‘T’ as one unit.

This unit with other 5 letters are to be arranged.

∴ The number of arrangement of one unit and 5 letters =

^{6}P

_{6}= 6! = 720

Also, ‘Y’ and ‘T’ can be arranged among themselves in

^{2}P

_{2}= 2! = 2 ways.

∴ A total number of arrangements when Y and T are always together = 6! × 2!

= 120 × 2

= 1440

(b) When ‘Y’ is next to ‘T’.

Let us take this (‘Y’ next to ‘T’) as one unit.

This unit with 5 other letters is to be arranged.

∴ The number of arrangements of 5 letters and one unit = ^{6}P_{6} = 6! = 720

Also, ‘Y’ has to be always next to ‘T’.

∴ They can be arranged among themselves in 1 way only.

∴ Total number of arrangements possible when Y is next to T = 720 × 1 = 720

(c) When there is no restriction.

7 letters can be arranged among themselves in ^{7}P_{7} = 7! ways.

∴ The total number of arrangements possible if there is no restriction = 7!

(d) When begin and end with a vowel.

There are 2 vowels in the word HISTORY.

All other letters of the word HISTORY are to be arranged between 2 vowels such that the arrangement begins and ends with a vowel.

The other 5 letters can be filled between the two vowels in ^{5}P_{5} = 5! = 120 ways.

Also, 2 vowels can be arranged among themselves at first and last places in ^{2}P_{2} = 2! = 2 ways.

∴ Total number of arrangements when the word begins and ends with vowel = 120 × 2 = 240

(e) When a word ends in ST.

As the arrangement ends with ST,

the remaining 5 letters can be arranged among themselves in ^{5}P_{5} = 5! = 120 ways.

∴ Total number of arrangements when the word ends with ST = 120

(f) When a word begins with S and ends with T.

As arrangement begins with S and ends with T,

the remaining 5 letters can be arranged between S and T among themselves in ^{5}P_{5} = 5! = 120 ways.

Total number of arrangements when the word begins with S and ends with T = 120

**Question 9.Find the number of arrangements of the letters in the word SOLAPUR so that consonants and vowels are placed alternately.Solution:**

There are 4 consonants S, L, P, R, and 3 vowels A, O, U in the word SOLAPUR.

Consonants and vowels are to be alternated.

∴ Vowels must occur in even places and consonants in odd places.

∴ 3 vowels can be arranged at 3 even places in

^{3}P

_{3}= 3! = 6 ways.

Also, 4 consonants can be arranged at 4 odd places in

^{4}P

_{4}= 4! = 24 ways.

Required number of arrangements = 6 × 24 = 144

**Question 10.Find the number of 4-digit numbers that can be formed using the digits 1, 2, 4, 5, 6, 8 if(a) digits can be repeated.(b) digits cannot be repeated.Solution:**

(a) A 4 digit number is to be made from the digits 1, 2, 4, 5, 6, 8 such that digits can be repeated.

∴ Unit’s place digit can be filled in 6 ways.

10’s place digit can be filled in 6 ways.

100’s place digit can be filled in 6 ways.

1000’s place digit can be filled in 6 ways.

∴ Total number of numbers that can be formed = 6 × 6 × 6 × 6 = 1296

(b) A 4 different digit number is to be made from the digits 1, 2, 4, 5, 6, 8 without repetition of digits.

∴ 4 different digits are to be arranged from 6 given digits which can be done in ^{6}P_{4} ways.

∴ Total number of numbers that can be formed

= 360

**Question 11.How many numbers can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition so that resulting numbers are between 100 and 1000?Solution:**
A number between 100 and 1000 that can be formed from the digits 0, 1, 2, 3, 4, 5 is of 3 digits, and repetition of digits is not allowed.

∴ 100’s place can be filled in 5 ways as it is a non-zero number.

10’s place digits can be filled in 5 ways.

Unit’s place digit can be filled in 4 ways.

∴ Total number of ways the number can be formed = 5 × 5 × 4 = 100

**Question 12.Find the number of 6-digit numbers using the digits 3, 4, 5, 6, 7, 8 without repetition. How many of these numbers are (a) divisible by 5 (b) not divisible by 5Solution:**

A number of 6 different digits is to be formed from the digits 3, 4, 5, 6, 7, 8 which can be done in

^{6}P

_{6}= 6! = 720 ways.

(a) If the number is to be divisible by 5,

the unit’s place digit can be 5 only.

∴ it can be arranged in 1 way only.

The other 5 digits can be arranged among themselves in

^{5}P

_{5}= 5! = 120 ways.

∴ Required number of numbers divisible by 5 = 1 × 120 = 120

(b) If the number is not divisible by 5,

unit’s place can be any digit from 3, 4, 6, 7, 8.

∴ it can be arranged in 5 ways.

Other 5 digits can be arranged in ^{5}P_{5} = 5! = 120 ways.

∴ Required number of numbers not divisible by 5 = 5 × 120 = 600

**Question 13.A code word is formed by two different English letters followed by two non-zero distinct digits. Find the number of such code words. Also, find the number of such code words that end with an even digit.Solution:**
There is a total of 26 alphabets.

A code word contains 2 English alphabets.

∴ 2 alphabets can be filled in

^{26}P

_{2}

= 650 ways.

Also, alphabets to be followed by two distinct non-zero digits from 1 to 9 which can be filled in

∴ Total number of a code words = 650 × 72 = 46800.

To find the number of codewords end with an even integer.

2 alphabets can be filled in 650 ways.

The digit in the unit’s place should be an even number between 1 to 9, which can be filled in 4 ways.

Also, 10’s place can be filled in 8 ways.

∴ Total number of codewords = 650 × 4 × 8 = 20800

**Question 14.Find the number of ways in which 5 letters can be posted in 3 post boxes if any number of letters can be posted in a post box.Solution:**
There are 5 letters and 3 post boxes and any number of letters can be posted in all three post boxes.

∴ Each letter can be posted in 3 ways.

∴ Total number of ways 5 letters can be posted = 3 × 3 × 3 × 3 × 3 = 243

**Question 15.Find the number of arranging 11 distinct objects taken 4 at a time so that a specified object (a) always occurs (b) never occursSolution:**

There are 11 distinct objects and 4 are to be taken at a time.

(a) The number of permutations of n distinct objects, taken r at a time, when one particular object will always occur is r×

^{ }

^{(n-1)}P(r−1)

∴ In 2880 permutations of 11 distinct objects, taken 4 at a time, one particular object will always occur.

(b) When one particular object will not occur, then 4 objects are to be arranged from 10 objects which can be done in ^{10}P_{4} = 10 × 9 × 8 × 7 = 5040 ways.

∴ In 5040 permutations of 11 distinct objects, taken 4 at a time, one particular object will never occur.

**Question 16.In how many ways can 5 different books be arranged on a shelf if(i) there are no restrictions(ii) 2 books are always together(iii) 2 books are never togetherSolution:**
(i) 5 books arranged in

^{5}P

_{5}= 5! = 120 ways.

(ii) 2 books are together.

Let us consider two books as one unit. This unit with the other 3 books can be arranged in ^{4}P_{4} = 4! = 24 ways.

Also, two books can be arranged among themselves in ^{2}P_{2} = 2 ways.

∴ Required number of arrangements = 24 × 2 = 48

**Question 17.3 boys and 3 girls are to sit in a row. How many ways can this be done if(i) there are no restrictions.(ii) there is a girl at each end.(iii) boys and girls are at alternate places.(iv) all-boys sit together.Solution:**

3 boys and 3 girls are to be arranged in a row.

(i) When there are no restrictions.

∴ Required number of arrangements = 6! = 720

(ii) When there is a girl at each end.

3 girls can be arranged at two ends in

And remaining 1 girl and 3 boys can be arranged between the two girls in ^{4}P_{4} = 4! = 24 ways.

∴ Required number of arrangements = 6 × 24 = 144

(iii) Boys and girls are at alternate places.

We can first arrange 3 girls among themselves in ^{3}P_{3} = 3! = 6 ways.

Let girls be denoted by G.

G – G – G –

There are 3 places marked by ‘-’ where 3 boys can be arranged in 3! = 6 ways.

∴ Total number of such arrangements = 6 × 6 = 36

OR

Similarly, we can first arrange 3 boys in 3! = 6 ways

and then arrange 3 girls alternately in 3! = 6 ways.

∴ Total number of such arrangements = 6 × 6 = 36

∴ Required number of arrangements = 36 + 36 = 72

(iv) All boys sit together.

Let us consider all boys as one group.

This one group with the other 3 girls can be arranged ^{4}P_{4} = 4! = 24 ways.

Also, 3 boys can be arranged among themselves in ^{3}P_{3} = 3! = 6 ways.

∴ Required number of arrangements = 24 × 6 = 144