Discussion

Explanation:

We have the following letters available:

A – 2, C – 1, I – 3, N – 1, O – 1, P – 2, R – 1, T – 2

Case 1: All 4 letters are different.

We have to select 4 letters out of 8 letters available. Number of ways = ⁸C₂ = 28.

These 4 letters can be arranged in 4! = 24 ways.

∴ Total number of such words = 28 × 24 = 672.

Case 2: 2 letters are same and other 2 are different.

2 same letters can be chosen from either A, I, O or P i.e., in 4 ways.

2 different letters can be chosen from the remaining 7 letters in ⁷C₂ = 21 ways.

∴ Total number of such selections = 4 × 21 = 84.

These 4 letters can be arranged in 4!/2! = 12 ways.

∴ Total number of such words = 84 × 12 = 1008.

Case 3: 2 same letters of 1 type and 2 same letters of other type

i.e., 2 pairs of letters can be chosen from either A, I, O or P in ⁴C₂ = 6 ways.

These 4 letters can be arranged in 4!/(2!×2!) = 6 ways.

∴ Total number of such words = 6 × 6 = 36.

Case 4: 3 letters are same and 1 other letter is different.

3 same letters can only be I i.e., 1 way.

1 other letter can be chosen from remaining 7 letters i.e., ⁷C₁ = 7 ways.

These 4 letters can be arranged in 4!/3! = 4 ways.

∴ Total number of such words = 4 × 7 = 28.

∴ Total number of ways 4 letter words can be formed = 672 + 1008 + 36 + 28 = 1744 ways.

Hence, 1744.

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