CAT 2025 Lesson : Permutations & Combinations - Conditional Permutations

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: ($3$) $^{10} \text{C}_{4} \times 5!$

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: $3600$

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 $1000$ to $3999$. 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: ($4$) $376$

A, B, C, D, E and F need to be seated in chairs numbered from $1$ to $6$. 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 $2$ or $5$, 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: $120$

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) $214$

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: $166$

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) $280$

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) $240$