MHT-CET Maths · Permutations and Combinations

Arrangements with Constraints — Together, Never Together, Fixed Positions and Repeated Letters

Arrange in a row under a condition: divide by k! for each letter repeated k times, glue a together-group into one block, place never-together items in the gaps, and fill a fixed position before counting the rest.

All Permutations and Combinations notes MHT-CET Maths strategy

Why this matters

10 PYQs at 50% HARD — the chapter's most expensive large page and the one where a single misread word costs the mark. Word stems recur with the same letters (CALCULATE, HAVANA, MANAMA, BARRACK) and the same three constraints — a fixed first and last letter, two letters kept apart, a group kept together — and the students-on-a-platform stem has been set twice. Every one is answered by the four moves below in some order; the difficulty is only in choosing the order.

Concept 1 of 4

Repeated Letters: Divide n! by k! for Each Repeat

Intuition

Swapping two identical A's changes nothing, so

n!

over-counts by the

k!

internal orders of each repeated letter. HAVANA has

6

letters with A three times:

\dfrac{6!}{3!} = 120

distinct words.

Definition

Arrangements of $n$ objects with $p$ alike, $q$ alike, …: $\dfrac{n!}{p!\,q!\cdots}$ .
Write the letter census FIRST. CALCULATE: C,C · L,L · A,A · U,T,E ( $9$ letters). MANAMA: M,M · A,A,A · N ( $6$ ). BARRACK: A,A · R,R · B,C,K ( $7$ ). 223355888: $2,2 \cdot 3,3 \cdot 5,5 \cdot 8,8,8$ .
Positions restricted by type: odd digits $3,3,5,5$ into the $4$ even positions in $\dfrac{4!}{2!\,2!} = 6$ ways, even digits $2,2,8,8,8$ into the $5$ odd positions in $\dfrac{5!}{2!\,3!} = 10$ ways: $60$ .
Short words from a multiset (four-letter words from BARRACK) go by cases on the repeat pattern: all different ${}^5P_4 = 120$ ; one pair + two singles ${}^2C_1 \cdot {}^4C_2 \cdot \dfrac{4!}{2!} = 144$ ; two pairs $\dfrac{4!}{2!\,2!} = 6$ ; total $270$ .

Permutations of a multiset

\frac{n!}{p!\,q!\,r!\cdots}

Worked example

How many distinct arrangements of the letters of BANANA have the two N's in the first and last positions?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Permutations and CombinationsMODERATE

The number of four-letter words that can be formed using letters of the word BARRACK is

[Q117 · 3rd May 2nd Shift · 2023]

Counting the letters wrong

Every word question is lost at the census. CALCULATE is nine letters, not eight; BARRACK has two R's AND two A's. Write the tally before any factorial.

Drill 1 more on repeated letters: divide n! by k! for each repeat

Concept 2 of 4

Fixed Positions: Fill the Constrained Slots First, Then the Rest

Intuition

A constraint on particular positions (first and last letter must be consonants;

B_1

sits second) removes choices, so those slots are filled first, and whatever is left fills the remaining slots freely.

Definition

CALCULATE starting and ending with a consonant: the consonants are C,C,L,L,T. Choose the ordered pair for the ends, then arrange the remaining $7$ letters with their repeats — summing over the end-pair patterns gives $\dfrac{5 \times 7!}{2}$ as the key has it.
Five students on a platform, $B_1$ in position $2$ , $G_1G_2$ adjacent: $B_1$ is placed ( $1$ way); the girls need two ADJACENT free positions from $\{1, 3, 4, 5\}$ — only $(3,4)$ and $(4,5)$ — in $2$ internal orders; the last two students fill the last two seats in $2!$ : $2 \times 2 \times 2 = 8$ .
Order of operations: most-constrained slot first. If two constraints compete (a fixed seat AND an adjacent pair), place the fixed one, then LIST the adjacent pairs that remain possible rather than assuming $n - 1$ of them.

Constrained slots first

\text{ways} = (\text{fill constrained slots}) \times (\text{arrange what remains})

Worked example

How many arrangements of the letters of ORANGE begin with a vowel and end with a consonant?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Permutations and CombinationsHARD

Five students are to be arranged on a platform such that the boy B1 occupies the second position and such that the girl G1 is always adjacent to the girl G2. Then, the number of such possible arrangements is

[Q113 · 11th May Shift 1 · 2024]

Assuming four adjacent pairs remain

After

B_1

takes seat

2

, seats

1

and

3

are no longer adjacent. The count of adjacent pairs is

2

, not

3

4

; option (B)

12

is what

3

pairs would give.

Drill 3 more on fixed positions: fill the constrained slots first, then the rest

Concept 3 of 4

Together: Glue the Group Into One Block, Then Arrange Inside It

Intuition

Items that must stay together are one object for the outer arrangement. Count the arrangement of the blocks, then multiply by the internal orders of each block.

Definition

$3$ Physics + $2$ Chemistry + $4$ Maths books, Physics together and Maths together: units are $P$ , $M$ , $C_1$ , $C_2$ — $4!$ ways; inside $P$ : $3!$ ; inside $M$ : $4!$ . Total $24 \times 6 \times 24 = 3456$ .
A block of $k$ distinct items contributes $k!$ internal orders; a block of identical items contributes $1$ .
'Exactly two letters repeated twice' in a $10$ -letter word from $10$ distinct letters: choose the two repeaters ${}^{10}C_2$ , the six singles ${}^8C_6$ , arrange $\dfrac{10!}{2!\,2!}$ ; dividing by the no-repeat count $10!$ gives $\dfrac{45 \times 28}{4} = 315$ .
The block method also proves the total for the complement: 'together' is the thing subtracted in every 'never together' count.

Block method

(\text{blocks})! \times \prod (\text{size of each block})!

Worked example

In how many ways can

4

boys and

3

girls stand in a row if all the girls stand together?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Permutations and CombinationsMODERATE

If 3 books on Physics, 2 books on Chemistry and 4 books on Mathematics are to be arranged on a shelf so that all the Physics books are together and all the Mathematics books are together, then the number of such arrangements is

[Q102 · 2nd May Shift 1 · 2023]

Forgetting the inside of the block

4! = 24

counts the blocks only; without

3! \times 4!

inside, the answer is

24

instead of

3456

. Conversely, a block of identical letters has NO inside factor.

Drill 1 more on together: glue the group into one block, then arrange inside it

Concept 4 of 4

Never Together: Total Minus Together, or Place Them in the Gaps

Intuition

Two ways to keep items apart. Complement: count all arrangements, subtract the ones with the items glued. Gap method: arrange everyone else first, then drop the restricted items into the

k + 1

gaps they create — never two in one gap.

Definition

Complement (HAVANA, V and N apart): total $\dfrac{6!}{3!} = 120$ ; V,N glued as a block $\dfrac{5!}{3!} \times 2 = 40$ ; apart $120 - 40 = 80$ .
Gap method (MANAMA, M's apart): arrange A,N,A,A in $\dfrac{4!}{3!} = 4$ ways; they make $5$ gaps; two IDENTICAL M's into two gaps: ${}^5C_2 = 10$ ; total $40$ . Distinct items in gaps use ${}^{k+1}P_r$ instead.
Choose the method by the count: two items apart — complement is quickest; three or more items pairwise apart — the gap method, since the complement needs inclusion-exclusion.
The gap method needs enough gaps: $r$ items apart need $r \le k + 1$ gaps, or the count is $0$ .

Two routes to 'apart'

\text{apart} = \text{total} - \text{together} \qquad \text{or} \qquad (\text{others})! \times {}^{k+1}P_r \ (\text{or } {}^{k+1}C_r \text{ if identical})

Worked example

In how many ways can

4

boys and

3

girls stand in a row so that no two girls are adjacent?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Permutations and CombinationsMODERATE

The number of ways of arranging letters of the word 'HAVANA', so that V and N do not appear together, is

[Q124 · May Shift 1 · 2021]

Using nPr for identical items in gaps

Two identical M's into

5

gaps is

{}^5C_2 = 10

, not

{}^5P_2 = 20

. The doubled answer

80

is option (C) on the MANAMA stem.

Drill 1 more on never together: total minus together, or place them in the gaps

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Repeated Letters: Divide n! by k! for Each Repeat
Permutations of a multiset
$\frac{n!}{p!\,q!\,r!\cdots}$
Fixed Positions: Fill the Constrained Slots First, Then the Rest
Constrained slots first
$\text{ways} = (\text{fill constrained slots}) \times (\text{arrange what remains})$
Together: Glue the Group Into One Block, Then Arrange Inside It
Block method
$(\text{blocks})! \times \prod (\text{size of each block})!$
Never Together: Total Minus Together, or Place Them in the Gaps
Two routes to 'apart'
$\text{apart} = \text{total} - \text{together} \qquad \text{or} \qquad (\text{others})! \times {}^{k+1}P_r \ (\text{or } {}^{k+1}C_r \text{ if identical})$

Watch out for (4)

Counting the letters wrong
→ Repeated Letters: Divide n! by k! for Each Repeat
Assuming four adjacent pairs remain
→ Fixed Positions: Fill the Constrained Slots First, Then the Rest
Forgetting the inside of the block
→ Together: Glue the Group Into One Block, Then Arrange Inside It
Using nPr for identical items in gaps
→ Never Together: Total Minus Together, or Place Them in the Gaps

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