+91 9600 121 800

Back

Quant

Modern Maths

Permutations & Combinations

ALL MODULES

CAT 2025 Lesson : Permutations & Combinations - Conditional Permutations

2.4 Conditional Permutations

When a permutation question has conditions or constraints, there are two commonly used approaches. The first approach is to first select the items and then arrange (Examples 18 and 19). This applies well where there is a constraint on either one of selection or arrangement. The other approach is to use blanks and apply the conditions (Example 20).

Example 18

In how many ways can a class teacher randomly pick

4

students from a group of

10

and sit along with them in a row?
(

1

)

^{10} \text{P}_{4}

(

2

)

^{10} \text{P}_{4} \times 5!

(

3

)

^{10} \text{C}_{4} \times 5!

(

4

)

^{10} \text{C}_{4} \times 4!

Solution

Number of ways to select

4

students =

^{10} \text{C}_{4}

The

4

selected students and the class teacher can be arranged in a row in

5!

ways.

\therefore

Total Permutations

=

^{10} \text{C}_{4} \times 5!

Answer: (

3

)

^{10} \text{C}_{4} \times 5!

Example 19

There are

5

Indians and

6

Chinese working in a team. The manager has to form a

4

-member project team with one member working on content, one on database and one each on front-end and back-end. Each of the 11 members are equally trained to take up any of the four mentioned parts of the project. If he needs to have two Chinese and two Indians in this team of 4, in how many ways can this team be formed?

Solution

2

Chinese and

2

Indians can be selected in

^{6} \text{C}_{2}

and

^{5} \text{C}_{2}

ways respectively. The

4

selected members can then be arranged across the

4

different roles in

^{4} \text{P}_{4}

=

4!

ways.

\therefore

Total Permutations =

^{6} \text{C}_{2} \times ^{5} \text{C}_{2} \times 4!

3600

Answer:

3600

Example 20

How many integers greater than

999

but not greater than

4000

, can be formed with the digits

0, 1, 2, 3

and

4

with repetition?
[CAT 2008]

(

1

)

499

(

2

)

500

(

3

)

375

(

4

)

376

(

5

)

501

Solution

Let's first consider the numbers from

1000

3999

. The thousands place can take

3

values (1, 2, 3), while the other three places can take all

5

digits as repetition is allowed. Number of Permutations =

\underline{3} \times \underline{5} \times \underline{5} \times \underline{5} = 375

It should be noted that

4000

should also be included as the condition states that the number should not be more than

4000

. Therefore, total permutations

=

375 + 1

=

\boldsymbol{376}

Answer: (

4

)

376

In examples 21,22 and 23, we are breaking into different cases and then adding.

Example 21

A, B, C, D, E and F need to be seated in chairs numbered from

1

6

. A can sit in a prime numbered seat only, while B has to sit in a seat that is a multiple of

3

. In how many ways can the

6

be seated?

Solution

Since the conditions are overlapping for the seat numbered

3

, which is prime as well as even, we need to consider two scenarios. This can be solved starting with A or B.
When A is in seat

3

, B has 1 seat to sit in (6) and the rest have

4

seats left.
Number of Permutations =

1 \times 1 \times 4!

24

When A is in seats

2

5

, B has

2

seats to sit in (3, 6) and the rest have

4

seats left.
Number of Permutations =

2 \times 2 \times 4!

96

Total Permutations =

24 + 96

120

Alternatively (Where B is seated first)

Where B is in seat

3

, A has

2

seats to sit in (2, 5) and the rest have

4

seats left.
Number of Permutations =

1 \times 2 \times 4!

48

Where B is in seat

6

, A has

3

seats to sit in (2, 3, 5) and the rest have

4

seats left.
Number of Permutations =

1 \times 3 \times 4!

72

Total Permutations =

48 + 72

120

Answer:

120

Example 22

How many whole numbers between

100

and

800

contain the number

2

?
[XAT 2013]

(1)

200

(2)

214

(3)

220

(4)

240

(5)

248

Solution

We look at the numbers from left to right. To avoid overlap, we start counting cases with

2

starting with the left-most digit (hundreds digit). Also, note that the hundreds place can contain

7

digits

(1, 2, 3, 4, 5, 6, 7)

while the tens and units places can have

10

digits each. The number

800

doesn't have

2

and can be ignored.

(1) Only the hundreds place has

\bm{2} =

\underline{1} \times \underline{10} \times \underline{10} = 100

, because the hundreds place can have only

2

and the tens and units places can have all the ten digits.

(2) Hundreds place has no

\bm{2}

and tens place has only

\bm{2} =

\underline{6} \times \underline{1} \times \underline{10} = 60

, because hundreds place can have

1, 3, 4, 5, 6

and

7

, tens place can have only

2

and units place can have all ten digits.

(3) Only units place has

\bm{2} =

\underline{6} \times \underline{9} \times \underline{1} = 54

, because hundreds place can take

6

values

(1, 3, 4, 5, 6

and

7)

, tens place can take all digits except

2

and units place can take only

2

.

Total number of ways in which the numbers can be formed are

100 + 60 + 54 = 214

Answer: (2)

214

Example 23

How many different

3

letter words (the words need not be in the dictionary) can be formed by using the letters in the word PARALLELED?

Solution

The letters arranged on the descending order of occurrence are L, L, L, A, A, E, E, P, R, D

As some of the letters are repeated, there are

6

unique letters – L, A, E, P, R, D.

We will have to select and arrange to find the number of

3

-letter words. However, depending on whether the letters in the

3

-letter word are repeated or not, different permutations are possible. There are three possible cases here.

1) All letters are the same
Only L occurs

3

times. Only

\bm{1}

word possible here – LLL.

2) Exactly

\bm{2}

letters are the same
L, A or E can form the repeated letter. Any of the remaining

5

letters after this is selected can form the

3^{rd}

letter.

Number of ways of selecting the repeated and non-repeated letter

= ^{3}C_{1} \times ^{5}C_{1} = 15

If two letters are repeated, number of permutations of the word

= \dfrac{3!}{2!} = 3

Total number of such words

= 15 \times 3 = \bm{45}

words

3) All 3 letters are distinct
Number of ways of selecting

3

letters from

6 = ^{6}C_{3} = 20

Number of words formed from using

3

distinct letters

= 3! = 6

Total number of distinct lettered words

= 20 \times 6 = \bm{120}

words

Total possible cases

= 1 + 45 + 120 = \bm{166}

Answer:

166

Example 24

There are

8

students of different heights. How many arrangements are possible when

4

students are picked and made to sit in a row,

(I) in the ascending order of their heights from left to right?
(II) where the tallest and the shortest of the

4

sit at the ends?

Solution

(I)

4

out of a total of

8

students can be selected in

^{8}C_{4}

ways.
The selected students can now be arranged in ascending order in only one way.

\therefore

Total Permutations

= ^{8}C_{4} \times 1 =70

(II) Once again,

4

out of

8

students can be selected in

^{8}C_{4}

ways.
Now, the tallest and shortest students can be arranged in either of the two ends in

2!

ways, while the remaining

2

students can be arranged in the remaining two places in

2!

ways.

\therefore

Total Permutations

= ^{8}C_{4} \times 2! \times 2! = 280

Answer: (I)

70

; (II)

280

Example 25

A team leader feels her

6

team members – A, B, C, D, E and F – have worked equally hard that year. She decides to randomly assign different ratings to them. In how many ways can the ratings be assigned such that
(I) A has a better rating than B?
(II) A has a better rating than B and C?

Solution

Let's look at this question as arranging the letters A, B, C, D, E, F in a word. There are

\bm{6! = 720}

ways

Let the order in which the word is written indicate the descending order of rating from left to right.

For instance, if the word is BDAFEC, then B's rating is the highest, followed by D, A, F, E and C.

(I) With A, there are two equally occurring possibilities that are A > B or B > A. Of the

\bm{2}

possible cases the desired case in these is the

\bm{1}

case where A > B.

\therefore

Permutations where A's rating is better than B's

= \dfrac{1}{2} \times 720 = \bm{360}

ways

(II) Ratings of A, B and C can occur in

3! = 6

ways (ABC, ACB, BAC, BCA, CAB, CBA).
Where in A > B > C or A > C > B, etc.
With no other bias or conditions, these

\bm{6}

possible cases are equally likely. The desired cases in these

6

are when A's rating is better than B and C, which is

\bm{2}

cases – A > B > C and A > C > B

\therefore

Permutations where A's rating is better than B's and C's

= \dfrac{2}{6} \times 720 = \bm{240}

ways

Answer: (I)

360

; (II)

240

Loading Video....