Ten different letters of alphabets are given. Words with five letters are formed from these given letters. Then, the number of words which have atleast one repeated letter is :

Correct — option (1), 69760.

A "word" here is any arrangement of 5 letters, meaningful or not. Letters may repeat, since the stem asks about words that have a repeated letter.

Step 1 — count all 5-letter words with repetition allowed. Each of the 5 places can take any of the 10 letters.
Total = 10 × 10 × 10 × 10 × 10 = 10⁵ = 100,000

Step 2 — count words in which no letter repeats. The five places have 10, 9, 8, 7 and 6 choices.
10 × 9 = 90; 90 × 8 = 720
720 × 7 = 5,040; 5,040 × 6 = 30,240

Step 3 — every other word has at least one repeated letter.
At least one repeat = 100,000 − 30,240 = 69,760

The idea to remember: "at least one" is counted as the total minus "none"; here, all words minus the words whose five letters are all different.

Why the others are wrong

• (2)
  30240 — 30,240 is the number of words with no repeated letter: ¹⁰P₅ = 10 × 9 × 8 × 7 × 6.

  It counts the opposite of what the stem asks. Subtracting it from all 100,000 words leaves the 69,760 words with at least one repeat.

• (3)
  99748 — 99,748 equals 100,000 − 252, and 252 is ¹⁰C₅, the number of ways to choose 5 of the 10 letters when order does not matter.

  A word is an arrangement, so order matters. The words with no repeated letter number ¹⁰P₅ = 30,240, not 252, which leaves 69,760.

• (4)
  99784 — 99,784 equals 100,000 − 216. That would mean just 216 words have all five letters different.

  The count of such words is 10 × 9 × 8 × 7 × 6 = 30,240, so the words with a repeated letter number 100,000 − 30,240 = 69,760.

The fundamental counting principle: if one step can be done in m ways and the next in n ways, the two together can be done in m × n ways.

Arranging r items in order from n different items gives nʳ arrangements when repetition is allowed, and ⁿPᵣ = n!/(n − r)! when it is not.

Choosing r items with order ignored gives ⁿCᵣ = n!/[r!(n − r)!]. Each such choice can be ordered in r! ways, so ⁿPᵣ = ⁿCᵣ × r!.

A count of "at least one" is found as the total minus the count of "none".

Key facts

• Arrangements of r items from n with repetition allowed: nʳ; here 10⁵ = 100,000.
• Arrangements of r items from n without repetition: ⁿPᵣ = n!/(n − r)!; here ¹⁰P₅ = 30,240.
• Selections of r items from n with order ignored: ⁿCᵣ = n!/[r!(n − r)!]; here ¹⁰C₅ = 252.
• ⁿPᵣ = ⁿCᵣ × r!; for n = 10 and r = 5, 252 × 120 = 30,240.
• Count of "at least one" = total count − count of "none".

Common traps

• Subtracting ¹⁰C₅ = 252 instead of ¹⁰P₅ = 30,240. A word is an ordered arrangement, so the no-repeat count uses permutations.
• Reading "at least one repeated letter" as "exactly one repeated letter". The count also includes words with two repeated letters or a letter used three or more times.
• Taking the total as ¹⁰P₅. With repetition allowed, every one of the 5 places has all 10 letters available, so the total is 10⁵.

UnlockIAS will link similar questions from RAS Pre 2023 and 2021 here once those papers are published on this site.

• practice — not a real PYQ

  How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 in which at least one digit is repeated?

  • (a)60
  • (b)65
  • (c)115
  • (d)125
  Answer(2) — All numbers: 5³ = 125; no repeat: 5 × 4 × 3 = 60; at least one repeat: 125 − 60 = 65. Option (1) counts numbers with no repeat; option (3) subtracts ⁵C₃ = 10 instead of 60; option (4) counts all numbers.
• practice — not a real PYQ

  In how many ways can three prizes of different values be given to 3 of 7 students, if no student gets more than one prize?

  • (a)35
  • (b)210
  • (c)343
  • (d)21
  Answer(2) — Order matters because the prizes differ: ⁷P₃ = 7 × 6 × 5 = 210. Option (1) is ⁷C₃, which ignores who gets which prize; option (3), 7³, lets one student win several prizes; option (4) is ⁷C₂.