Permutations When Objects Are Not All Distinct

Why ROOT Has 12 Arrangements, Not 24

Rearrange the letters of ROOT. If the two O's were distinguishable — $O_1$ and $O_2$ — there would be $4! = 24$ arrangements. But $RO_1O_2T$ and $RO_2O_1T$ are the same word ROOT. Every genuine word is counted $2!$ times, once per internal ordering of the O's.

Key Point (Theorems 3-4): The number of permutations of $n$ objects where $p_1$ are alike of one kind, $p_2$ alike of a second kind, …, $p_k$ alike of a $k$-th kind (rest all different) is

$$\frac{n!}{p_1!\,p_2!\cdots p_k!}$$

One factorial in the denominator per repeated kind. INSTITUTE (2 I's, 3 T's): $\frac{9!}{2!\,3!} = 30240$. ALLAHABAD (4 A's, 2 L's): $\frac{9!}{4!\,2!} = 7560$.

[Board Tip] First write the letter inventory: total count, then each repeated letter with its multiplicity. MISSISSIPPI = 11 letters: 4 I's, 4 S's, 2 P's $\Rightarrow \frac{11!}{4!\,4!\,2!} = 34650$. The inventory line earns method marks even if arithmetic slips.

Digit-Building and Restrictions

The same machinery drives number-building problems:

• 4-digit numbers from digits 1-9, no repetition: order matters, digits distinct — $^9P_4 = 3024$.
• Numbers between 100 and 1000 from 0-5, no repetition: count all 3-digit strings $^6P_3 = 120$, then subtract those starting with 0 (really 2-digit numbers): fixing 0 in the hundreds place leaves $^5P_2 = 20$. Answer: $120 - 20 = 100$.

Key Point (the zero trap): When 0 is among the digits, strings with a leading 0 are not genuine numbers. Either lock the first place to a nonzero digit, or count all strings and subtract the leading-zero ones.

Restriction problems on words follow fixed patterns:

• Fix a position (words starting with P): place the fixed letter, arrange the rest — INDEPENDENCE starting with P: $\frac{11!}{3!\,4!\,2!} = 138600$.
• Vowels together: bundle the vowels into one block, arrange blocks, then arrange within — DAUGHTER: $6! \times 3! = 4320$.
• Vowels NOT (all) together: complement — all arrangements minus the together count: $8! - 4320 = 36000$.

[JEE Tip] In "never together" problems, be precise about the claim: "not all together" is the complement of one bundle; "no two adjacent" needs the stronger gap method (Section 6). The MISSISSIPPI four-I's problem below means not all four in one block.

Solved Examples

Example 1: ALLAHABAD

Find the number of permutations of the letters of ALLAHABAD.

Solution:

Step 1 — Write the letter inventory. 9 letters — 4 A's, 2 L's, the rest all different.

Step 2 — Apply the repeated-objects formula. $\frac{9!}{4!\,2!}$ — one factorial per repeated kind.

Step 3 — Evaluate. $\frac{362880}{24 \times 2} = \frac{362880}{48} = 7560$.

Takeaway: The inventory line comes first, always — it dictates the entire denominator.

Example 2: 4-digit numbers

How many 4-digit numbers can be formed using digits 1 to 9, repetition not allowed?

Solution:

Step 1 — Identify the setting. Order matters, all nine digits distinct, no zero to worry about: a straight permutation count.

Step 2 — Apply the formula. $^9P_4 = 9 \times 8 \times 7 \times 6 = 3024$.

Takeaway: With no 0 in the digit set, number-building is pure $^nP_r$.

Example 3: The zero trap

How many numbers between 100 and 1000 can be formed with digits 0, 1, 2, 3, 4, 5, repetition not allowed?

Solution:

Step 1 — Rephrase the range. Numbers between 100 and 1000 are exactly the 3-digit numbers.

Step 2 — Count all 3-digit strings. $^6P_3 = 6 \times 5 \times 4 = 120$.

Step 3 — Subtract the fakes. Strings starting with 0 are really 2-digit numbers: fix 0 first, arrange 2 of the remaining 5: $^5P_2 = 20$.

Step 4 — Conclude. $120 - 20 = 100$.

Takeaway: When 0 is in the digit set, either lock the first place to nonzero or count-and-subtract the leading-zero strings.

Example 4: Digit drills

(i) How many 3-digit numbers from digits 1-9, no repetition? (ii) How many 4-digit numbers exist in all (digits 0-9, no repetition)?

Solution:

Step 1 — (i) No zero present. $^9P_3 = 9 \times 8 \times 7 = 504$.

Step 2 — (ii) Zero present: lock the first place. First digit nonzero: 9 ways; then 9, 8, 7 remaining digits for the other places.

Step 3 — Multiply. $9 \times 9 \times 8 \times 7 = 4536$.

Takeaway: The same problem with and without 0 — the leading-place lock is the only difference.

Example 5: 4-digit and even

From digits 1-5 (no repetition): how many 4-digit numbers, and how many of them are even?

Solution:

Step 1 — All 4-digit numbers. $^5P_4 = 5 \times 4 \times 3 \times 2 = 120$.

Step 2 — Even: lock the units place first. Units must be 2 or 4: 2 ways.

Step 3 — Fill the rest. Three places from the remaining four digits: $^4P_3 = 24$.

Step 4 — Multiply. $2 \times 24 = 48$.

Takeaway: Parity restrictions live in the units place — lock it first, then permute the rest.

Example 6: DAUGHTER vowels

How many 8-letter arrangements of DAUGHTER have (i) all vowels together (ii) vowels not all together?

Solution:

Step 1 — (i) Bundle the vowels. Tie A, U, E into one block; with the 5 consonants that makes 6 objects: $6!$ arrangements.

Step 2 — Arrange inside the bundle. The three vowels order themselves in $3!$ ways.

Step 3 — Multiply. $6! \times 3! = 720 \times 6 = 4320$.

Step 4 — (ii) Complement. All arrangements minus the together count: $8! - 4320 = 40320 - 4320 = 36000$.

Takeaway: Bundle for "together"; complement for "not all together".

Example 7: Coloured discs

In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row (same-colour discs indistinguishable)?

Solution:

Step 1 — Inventory. $n = 9$ discs in kinds of sizes 4, 3, 2.

Step 2 — Apply the formula. $\frac{9!}{4!\,3!\,2!} = \frac{362880}{288} = 1260$.

Takeaway: Identical objects of each colour behave exactly like repeated letters.

Example 8: INDEPENDENCE

For the word INDEPENDENCE (12 letters: 3 N's, 4 E's, 2 D's), count the arrangements in which (i) all (ii) words start with P (iii) all vowels occur together (iv) vowels never all together (v) words begin with I and end in P.

Solution:

Step 1 — (i) Total. $\frac{12!}{3!\,4!\,2!} = 1663200$.

Step 2 — (ii) Fix P at the left. The remaining 11 letters keep their repeats: $\frac{11!}{3!\,4!\,2!} = 138600$.

Step 3 — (iii) Bundle the vowels EEEEI. Block + 7 consonants = 8 objects with 3 N's and 2 D's: $\frac{8!}{3!\,2!} = 3360$; inside the block the vowels arrange in $\frac{5!}{4!} = 5$ ways. Total $3360 \times 5 = 16800$.

Step 4 — (iv) Complement. $1663200 - 16800 = 1646400$.

Step 5 — (v) Pin both ends. I left, P right; the middle 10 letters: $\frac{10!}{3!\,4!\,2!} = 12600$.

Takeaway: One word, five standard restriction patterns — fixing, bundling, complementing and end-pinning all in one drill.

Example 9: EQUATION and MONDAY

(i) How many words use all letters of EQUATION exactly once? (ii) From MONDAY (no repetition): words using 4 letters; all 6 letters; all letters with a vowel first?

Solution:

Step 1 — (i) All distinct. EQUATION has 8 distinct letters: $8! = 40320$.

Step 2 — (ii) Four at a time. $^6P_4 = 6 \times 5 \times 4 \times 3 = 360$.

Step 3 — All six. $6! = 720$.

Step 4 — Vowel first. First place O or A (2 ways), the rest in $5!$: $2 \times 120 = 240$.

Takeaway: Distinct-letter words are plain factorials and falling products — the repeats machinery switches off.

Example 10: MISSISSIPPI

In how many distinct permutations of MISSISSIPPI do the four I's NOT come together?

Solution:

Step 1 — Total arrangements. Inventory: 11 letters — 4 I's, 4 S's, 2 P's: $\frac{11!}{4!\,4!\,2!} = 34650$.

Step 2 — Count the four-I's-together arrangements. Bundle IIII (internally identical — 1 way); with the 7 other letters that is 8 objects with 4 S's and 2 P's: $\frac{8!}{4!\,2!} = 840$.

Step 3 — Complement. $34650 - 840 = 33810$.

Takeaway: "Not together" for a block of identical letters = total minus one-block count; the identical letters contribute no internal factor.

Example 11: PERMUTATIONS

In how many ways can the letters of PERMUTATIONS be arranged if (i) words start with P and end with S (ii) vowels are all together (iii) there are always 4 letters between P and S?

Solution:

Step 1 — Inventory. 12 letters with T appearing twice.

Step 2 — (i) Pin both ends. P first, S last; the middle 10 letters (2 T's): $\frac{10!}{2!} = 1814400$.

Step 3 — (ii) Bundle the vowels. E, U, A, I, O are all distinct — one block + 7 consonants (2 T's) = 8 units: $\frac{8!}{2!} = 20160$; vowels inside: $5!= 120$. Total $20160 \times 120 = 2419200$.

Step 4 — (iii) Slide the (P, S) frame. Exactly 4 letters between them means positions $(i, i+5)$ for $i = 1, \ldots, 7$ — 7 slots, times 2 for swapping P and S: 14 ways. The other 10 letters fill the rest: $\frac{10!}{2!} = 1814400$.

Step 5 — Multiply. $14 \times 1814400 = 25401600$.

Takeaway: Distance conditions become sliding-frame counts: enumerate the frame positions, then permute the rest.