CAT 2025 Lesson : Permutations & Combinations - Special Types

In how many ways can all the letters in the word FINANCE be arranged such that
(I) the two N's are never together
(II) the vowels occupy even places?
(III) the vowels occupy odd places?
(IV) all consonants are together
(V) the vowels are never together?

Solution

Case I: Total number of words = $\dfrac{7!}{2!}$ = $2520$

From this we can subtract the cases where Ns are always together. Here, we can treat the two Ns as 1 letter and permute the letters of the word. Now, we have F, I, A, C, E and NN (which is considered to be 1 letter).

Words where Ns are together = $6!$ = $720$

Number of words where Ns are never together = $2520 - 720$ = $\boldsymbol{1800}$

Case II: The $3$ vowels (I, A, E) can occupy $3$ even places (2nd, 4th and 6th letters) in $3!$ = $6$ ways.

The remaining letters (F, N, N, C) can occupy the $4$ odd places in $\dfrac{4!}{2!}$ = $12$ ways.

$\therefore$ Total Permutations = $6 \times 12$ = $72$

Case III: Condition is on $3$ vowels occupying $3$ of the $4$ odd places. We can select the $3$ odd places for the vowels in $^{4} \text{C}_{3}$ ways and then permute the vowels in $3!$ ways. We can then assign the remaining letters (F, N, N, C) in $\dfrac{4!}{2!}$ ways.
$\therefore$ Total Permutations = $^{4} \text{C}_{3} \times 3! \times \dfrac{4!}{2!}$ $=$ $4 \times 6 \times 12$ $= \boldsymbol{288}$

Case IV: The consonants F, N, N and C can be treated as a single letter or block. Now, we have 4 letters/blocks – I, A, E, (FNNC). These can be permuted in 4! = 24 ways.

Within the block, F, N, N, C can be permuted in $\dfrac{4!}{2!}$ = $12$ ways.

$\therefore$ Total Permutations = $24 \times 12$ = $\boldsymbol{288}$

Case V: When the vowels are never together, we can use the slotting technique (covered in Section 3.1).

Lets place the consonants (F, N, N, C) first ⇒ $\dfrac{4!}{2!}$ = $12$ ways

There are $5$ slots between these consonants ( _F_N_N_C_ ), where the vowels can be slotted, so that they are not next to each other. We can now select the $3$ slots and then arrange the vowels.

Number of ways to select $3$ from $5$ and arrange $3$ vowels = $^{5} \text{C}_{3} \times 3!$ = $60$ ways.

$\therefore$ Total Permutations = $12 \times 60$ = $\boldsymbol{720}$

Answer: (I) $1800$; (II) $72$; (III) $288$; (IV) $288$; (V) $720$

A group of $7$ students need to be seated in a row. Two students had a quarrel and, therefore, cannot be seated next to each other. In how many ways can the students be seated?

Solution

Permutations for $7$ students can be seated = $7!$ = $5040$

If the $2$ students are to be always together, they can be treated as one unit. So, $6$ students/units to be permuted overall and within $1$ one of the units, $2$ of them to be permuted.
Permutations where the $2$ are always together = = 1440

Permutations where the $2$ are never together = $5040 - 1440$ = $\boldsymbol{3600}$

Alternatively

To slot the $2$ quarrelling students, we first seat the $5$ other students in $5!$ ways. We now have $6$ slots between these $5$ students and at the ends as shown below (where X represents a student seated and __ represents an empty slot).

__ X __ X __ X __ X __ X __

We can select these slots in $^{5} \text{C}_{2}$ ways and permute the quarrelling students in $2!$ ways.

$\therefore$ Permutations where the $2$ are never together = $5! \times ^{6} \text{C}_{2} \times 2!$ = $120 \times 15 \times 2$ = $\boldsymbol{3600}$

Answer: $3600$

If all the letters in the word MATTER are permuted and arranged in alphabetical order, what is the rank of MATTER?

Solution

In these questions, we find the number of words that alphabetically occur before MATTER. Note that the only letter that is repeated is T, which occurs twice.

Words that start with	Letters to be Permuted	Number of Words
A	MTTER	$\dfrac{5!}{2!}$ = $60$
E	MATTR	$\dfrac{5!}{2!}$ = $60$
MAE	TTR	$\dfrac{3!}{2!}$ = $3$
MAR	TTE	$\dfrac{3!}{2!}$ = $3$
MATE	TE	$2!$ = $2$
MATR	TE	$2!$ = $2$
MATTER	-	$0!$ = $1$

Rank = $60 + 60 + 3 + 3 + 2 + 2 + 1$ = $\boldsymbol{131}$

Answer: $131$

Note: As we are finding the rank, we need to add the 1 way of writing MATTER as well.

What is the sum of all $4$-digit numbers that can be formed with $3, 4, 5$ and $6$, where none of the digits are repeated?

Solution

Total numbers $= 4! = 24$
Each of these digits will appear equal number of times in each of the digit places – units, tens, hundreds and thousands places.

Therefore, each of the units digits will appear equal number of times, i.e. $\dfrac{24}{4} = 6$ times in each of the places.

Sum of digits in the units place $= (6 \times 3) + (6 \times 4) + (6 \times 5) + (6 \times 6)$ = $6 \times (3 + 4+ 5+ 6) = 108$

Therefore, in each of the digit places, the sum of digits (face values) will be $108$.

Sum of all the tens places $= 108 \times 10 = 1080$
Sum of all the hundreds places $= 108 \times 100 = 10800$
Sum of all the thousands places $= 108 \times 1000 = 108000$

Total value of the number $= 108000 + 10800 + 1080 + 108 = 119988$

Alternatively (Recommended) Sum of all the possible numbers $= (4 - 1)! \times (3 + 4 + 5 + 6) \times 1111 = 119988$

Answer: $119988$

What is the sum of all $5$-digit numbers that can be formed with $3, 4, 5, 6$ and $0$, where none of the digits are repeated?

Solution

Sum of all possible $5$-digit numbers including $0 = (5 - 1)! \times (3 + 4 +5 + 6 + 0) \times 11111 = 4799952$

However, it is to be noted that when $0$ occupies the ten-thousands place, the number becomes a $4$-digit numbers. These numbers have also been added in the above result of $4799952$.

Therefore, to eliminate these, we need to subtract the total of all numbers where $0$ is in ten-thousands place. This is nothing but the $4$-digit numbers that can be formed using $3, 4, 5$ and $6$.

Sum of all possible $4$-digit numbers without $0 = (4 - 1)! \times (3 + 4 + 5 + 6) \times 1111 = 119988$

Sum of all $5$-digit numbers excluding $0$ in ten-thousands place $= 4799952 - 119988 = 4679964$

Answer: $4679964$

A person can either take a step forward or back. After taking $9$ steps he has to be exactly $3$ steps ahead of the point where he started. In how many ways can this be achieved?

Solution

To move exactly $3$ steps ahead after taking $9$ steps, one must take 6 forward steps and $3$ backward steps. This can be in any order. Let the forward step be denoted by F and backward step be denoted by B.

FFFFFFBBB denotes the person has taken $6$ forward steps to begin with and then has taken $3$ backward steps.
However, other combinations like FFBBBFFFF or FBFBFBFFF are also possible.

Number of ways in which FFFFFFBBB can be permuted is the number of ways we need to find.

Permutations of the word FFFFFFBBB $= \dfrac{9!}{6! \times 3!} =$ $\dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$

Answer: $84$

In the image below, the vertical and horizontal lines represent the roads in a city. At points of intersection, one can move from one road to the other. Mary has to travel from points A to B taking the shortest possible path.
(I) In how many ways can she travel from A to B?
(II) In how many ways can she travel from A to B if she has to pass through point C, where she picks up her kid from school?

Solution

Case I: Horizontal lines are parallel to one another. Likewise, the vertical lines are also parallel.

Let a movement upwards be denoted by U and the movement rightwards be denoted by R.

The shortest route, irrespective of the order, comprises of $5$ Us and $5$ Rs. One such path is UUUUURRRRR.

UUUUURRRRR can be permuted in $\dfrac{10!}{5! \times 5!} = \bm{252}$ ways

Case II: The shortest path from A to C should have $2$ Us and $2$ Rs

UURR can be permuted in $\dfrac{4!}{2! \times 2!} = 6$ ways

The shortest path from C to B should have $3$ Us and $3$ Rs.

UUURRR can be permuted in $\dfrac{6!}{3! \times 3!} = 20$ ways

Shortest way to travel from A to C and then from C to B $= 6 \times 20 = 120$ ways

Answer: (I) $252$; (II) $120$

Note: The word and highlighted above clearly requires us to multiply the ways. Note that for every $1$ way from A to C, there are $20$ ways from C to B. As there are $6$ ways from A to C, one can go in $6 \times 20 = 120$ ways from A to B via C.