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Words with repeated letters are a JEE favourite because one formula answers the plain question and then a small condition, such as "begins and ends with S", catches anyone who applies the formula without thinking. This post explains where the formula comes from, works MISSISSIPPI under three conditions, and shows the case method for the harder question where you use only some of the letters.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists permutations and combinations with "Simple applications", and the JEE (Advanced) 2026 brochure lists permutations and combinations and uses them again in probability. The method is section 6.3.4, "Permutations when all the objects are not distinct objects", in the NCERT Class 11 chapter Permutations and Combinations (2026-27 reprint). Check the documents for your exam year.
Why do you divide by the factorials?
NCERT explains it with INSTITUTE, which has two I's and three T's. Pretend the repeated letters are different: I₁, I₂, T₁, T₂, T₃. Then there are 9! arrangements. But in any one word, the two I's can swap in 2! ways and the three T's in 3! ways without changing what you read. So each real word was counted 2! × 3! times, and the true count is 9!/(2! 3!).
NCERT then states the general result as Theorem 4: n objects, of which p₁ are of one kind, p₂ of a second kind and so on, can be arranged in n!/(p₁! p₂! … pₖ!) ways.
Check it on something small enough to list. SEE has three letters with two E's, so the formula gives 3!/2! = 3. The list is SEE, ESE, EES. Listing like this is not wasted time. In a study of undergraduates, Lockwood and Gibson found that even partial lists of the outcomes being counted were linked to better success on counting problems (ERIC abstract).
Worked example 1: MISSISSIPPI with no conditions
Count the letters first and write them down:
M: 1
I: 4
S: 4
P: 2
Total: 11
Arrangements = 11!/(4! × 4! × 2!) = 39,916,800/(24 × 24 × 2) = 39,916,800/1152 = 34,650.
The 1! for M is left out because it equals 1.
Worked example 2: begins and ends with S
Fix an S at each end. That uses two of the four S's. The middle nine places hold M, four I's, two S's and two P's:
9!/(4! × 2! × 2!) = 362,880/96 = 3,780.
The common slip is to keep dividing by 4! for the S's. Once two S's are fixed, only two are left to arrange, so the divisor is 2!. Recount the letters after every restriction.
Worked example 3: all four I's together
Glue the I's into one block, IIII. Now arrange 8 objects: M, the block, four S's and two P's.
8!/(4! × 2!) = 40,320/48 = 840.
Inside the block the I's are identical, so the block has only one internal order. With different letters you would multiply by the number of ways to order them inside the block; here that number is 1.
Exercise 6.3 of NCERT asks how many arrangements do not have the four I's together. Subtract: 34,650 − 840 = 33,810. "Not all together" is not the same as "no two together"; that second condition needs the gap method, which our post on arranging so that no two are together works through, including MISSISSIPPI.
Worked example 4: four-letter words from MISSISSIPPI
How many four-letter arrangements can be made from the letters of MISSISSIPPI? Now you cannot use one formula, because the answer depends on which letters repeat in the word. The available stock is M (1), I (4), S (4) and P (2). Split by pattern:
All four different (M, I, S, P): 4! = 24.
One pair and two different letters. The pair can be I, S or P: 3 ways. The two other letters come from the remaining three: C(3, 2) = 3. Each choice gives 4!/2! = 12 orders. Total 3 × 3 × 12 = 108.
Two pairs. Choose two of I, S, P: C(3, 2) = 3. Each gives 4!/(2! 2!) = 6. Total 18.
Three alike and one different. The triple is I or S: 2 ways. The single letter is any of the other three: 3 ways. Each gives 4!/3! = 4. Total 24.
Four alike. IIII or SSSS: 2.
Sum: 24 + 108 + 18 + 24 + 2 = 176. A brute-force count by computer gives the same 176. Notice that the pattern "three alike" excludes P, which appears only twice. Checking the stock for each case is the step that catches this.
Which mistakes cost marks?
Forgetting a repeated letter. Write the tally of every letter before using the formula.
Using the original divisor after fixing letters, as in example 2.
Multiplying a block of identical letters by a factorial.
Treating "not all together" and "no two together" as the same condition.
Using one formula when only some letters are chosen, instead of cases.
A practice routine
Tally and count BANANA, STATISTICS and ASSASSINATION. Check one answer by listing a smaller word with the same pattern.
Solve NCERT's Example 16 on INDEPENDENCE, then compare it with the gap method.
Do two "only some letters" problems by cases, and add up the cases twice.
For more on avoiding double counts, read our post on counting with permutations and combinations.
When a count comes out too big and you cannot find the double count, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Today, find the number of arrangements of MISSISSIPPI that begin with P, and check it is 34,650 × 2/11.
