How many different words, with or without meaning, can be formed by using the letters of the word COVID?

Q: What is the answer to CAPF 2021 Paper I Q80?

D. Correct — D, 120. COVID has five letters — C, O, V, I and D — and every one of them is different. Arranging five distinct objects in a row can be done in 5! ways, that is 5 × 4 × 3 × 2 × 1 = 120. The phrase 'with or without meaning' is the examiner's way of saying that every arrangement counts, so no dictionary check is required. Had a letter repeated, the count would have to be divided by the factorial of each repetition — the word LEVEL, for instance, gives 5! ÷ (2! × 2!) = 30 — but no such division applies here. (Provisional — pending the official UPSC key.)

(a)60
(b)150
(c)100
(d)120

Answer

Why

Correct — D, 120. COVID has five letters — C, O, V, I and D — and every one of them is different. Arranging five distinct objects in a row can be done in 5! ways, that is 5 × 4 × 3 × 2 × 1 = 120. The phrase 'with or without meaning' is the examiner's way of saying that every arrangement counts, so no dictionary check is required. Had a letter repeated, the count would have to be divided by the factorial of each repetition — the word LEVEL, for instance, gives 5! ÷ (2! × 2!) = 30 — but no such division applies here.

Why the others are wrong

(a)60 — Half of 120. It is what you get by dividing by 2! for a repeated letter that COVID does not contain.
(b)150 — Not a factorial or any quotient of one. It comes from mixing addition into the counting rather than multiplying the choices.
(c)100 — Suggests five choices for each of two positions, which would be the count if letters could repeat. Here each letter is used exactly once.

Concept

Arranging n distinct objects in a line gives n! orders, since there are n choices for the first position, n − 1 for the second and so on. When some objects are identical the arrangements are overcounted, and the correction is to divide by the factorial of the count of each repeated object. Choosing r of n objects and arranging them gives the permutation nPr = n!/(n − r)!, while choosing without arranging gives the combination nCr.

Letter-arrangement questions almost always come down to two checks. Count the letters, and look for repeats. In COVID both checks are trivial, which makes it a gift item worth taking quickly. The variants that follow are worth rehearsing: arrangements with the vowels together are counted by tying the vowels into one block, so COVID gives 4! × 2! = 48; arrangements beginning with a particular letter fix that position and permute the rest, giving 4! = 24.

Key facts

COVID has five letters and no repetition, so the count is 5! = 120.
'With or without meaning' means every arrangement is admissible.
Repeated letters require division by the factorial of each repetition.
Tying the two vowels of COVID together gives 4! × 2! = 48 arrangements.
Fixing one letter in the first position leaves 4! = 24 arrangements.

Five positions, five distinct letters

First position→5 choices

Second position→4 choices remain

Third, fourth, fifth→3, then 2, then 1

Total→5 × 4 × 3 × 2 × 1 = 120

If a letter repeated twice→Divide by 2!, giving 60 — the value offered as a distractor

The only real check is for repeated letters, and COVID has none.

Study next

Permutations of distinct and of repeated objectsPermutations with restrictions — letters together, letters apart, fixed positionsCombinations, and when order does not matterCircular permutations, where one object is fixed to break the rotationThe addition and multiplication principles of counting

Common traps

Dividing by a factorial for a repetition that is not there.
Counting only meaningful words when the stem says with or without meaning.
Using 5⁵ or 5², which would apply if letters could be reused.

The simplest permutation item there is. Count the letters, check for repeats, take the factorial.

Related PYQs

UPSC_2010_GS1_Q1172010Open & attempt →

In how many ways can four children be made to stand in a line such that two of them, A and B, are always together?

(a) 6
(b) 12
(c) 18
(d) 24

Answer(b) 12

The block method on a smaller set. Treating the pair as one unit gives 3! orders and 2! internal ones, and the unrestricted count for four distinct children would be 4! = 24.

CAPF_GAI_2022_Q412022Open & attempt →

There is a group of 5 people among which there is one couple. In how many ways can these 5 people be seated in a row having 5 chairs if the couple is to be seated next to each other?

(a) 24
(b) 48
(c) 60
(d) 120

Answer(b) 48

The same five-object arrangement with one restriction added. Tying the couple together leaves four units to arrange in 4! ways and two orders within the pair, giving 48 out of the unrestricted 120 that this item counts.

Practice

practice — not a real PYQ
How many different arrangements can be made of the letters of the word DELHI?
- (a)24
- (b)60
- (c)120
- (d)720
Answer(c) 120 — five distinct letters, so 5!.
practice — not a real PYQ
How many arrangements can be made of the letters of the word LEVEL?
- (a)20
- (b)30
- (c)60
- (d)120
Answer(b) 30 — 5! divided by 2! for the two L's and 2! for the two E's.