Selecting 3 letters from 5 and arranging them: ⁵P₃ = 60. Order matters here.

How Many 3-Letter Words from GIVEN? — Combination vs Permutation

Question

How many 3-letter words can be formed using the letters of the word GIVEN?

(NCERT Class 11, Chapter 7 — Exercise 7.3)

Solution — Step by Step

The word GIVEN has 5 letters: G, I, V, E, N. All five are distinct — no repeats. This matters because repeated letters would need a different approach.

We’re forming words, meaning the order of letters matters. The word “GIV” is different from “VIG”. Whenever order matters, we use permutation, not combination.

We’re choosing 3 letters from 5 and arranging them. The formula is:

^nP_r = \frac{n!}{(n-r)!}

Plugging in $n = 5$ , $r = 3$ :

^5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60

Fill 3 positions one at a time:

1st letter: 5 choices
2nd letter: 4 remaining choices
3rd letter: 3 remaining choices

5 \times 4 \times 3 = 60 \checkmark

Answer: 60 three-letter words can be formed.

Why This Works

The slot method gives us the clearest picture of what’s happening. Each time we place a letter, the pool shrinks by one — because we can’t reuse the same letter (the problem doesn’t say “with repetition allowed”).

The permutation formula $^nP_r$ is just this multiplication written compactly. When we write $\frac{5!}{2!}$ , we’re computing $5 \times 4 \times 3 \times \underbrace{2 \times 1}_{\text{these cancel}} / \underbrace{2 \times 1}_{2!}$ . The $2!$ in the denominator cancels out exactly the part we don’t need.

This type of question — “how many arrangements of $r$ things from $n$ distinct things” — is a standard 2-3 mark question in both CBSE Class 11 and JEE Main. The word “GIVEN” or similar 5-letter words with distinct letters appear repeatedly across NCERT exercises.

Alternative Method

Using $^nC_r$ then arranging:

Some students find it cleaner to split into two steps:

Select 3 letters from 5: $^5C_3 = \frac{5!}{3! \cdot 2!} = 10$ ways
Arrange those 3 letters: $3! = 6$ ways

Total = $10 \times 6 = \mathbf{60}$

This approach makes the logic very explicit — first choose which letters, then decide order. Both methods give the same answer because $^nP_r = ^nC_r \times r!$ by definition.

When a problem says “words” or “arrangements”, that’s your signal to use permutation. When it says “groups”, “teams”, “committees”, or “selections”, that’s combination. The question is always: does swapping the order give a different outcome?

Common Mistake

Using $^5C_3 = 10$ as the final answer.

Students confuse “selecting 3 letters” with “forming 3-letter words.” If the question asked “how many ways can you choose 3 letters from GIVEN,” then 10 would be correct. But “words” means ordered arrangements — so each selection of 3 letters generates $3! = 6$ different words. Missing this step loses you the full marks.

A quick check: if you had just 2 distinct letters, say A and B, how many 2-letter words can you form? Clearly 2 — “AB” and “BA”. $^2C_2 = 1$ , but $^2P_2 = 2$ . The permutation answer matches our intuition, so use that.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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