Question

If four letters are chosen from the letters of the word ASSIGNMENT and are arranged in all possible ways to form 4 letter words (with or without meaning), then total number of such words that can be formed is

• A. \( 1680 \)
• B. \( 2184 \)
• C. \( 2196 \)
• D. \( 2190 \) Correct Answer

Hint:

When forming words from a given set of letters with repetitions, categorize the selections of letters first: 1. All distinct letters. 2. Some letters alike, others distinct. 3. Multiple sets of alike letters. For each category, calculate the number of ways to select the letters, and then the number of ways to arrange them. The number of permutations of \(n\) objects where there are \(n_1\) identical objects of type 1, \(n_2\) of type 2, ..., \(n_k\) of type k is \( \frac{n!}{n_1! n_2! \dots n_k!} \).

Answer 1

The Correct Option is D

Step 1: Analyze the letters in the word ASSIGNMENT.
The letters are A, S, S, I, G, N, M, E, N, T.
(Re-checking: A, S, S, I, G, N, N, M, E, T) Frequencies of letters: A: 1 S: 2 I: 1 G: 1 N: 2 M: 1 E: 1 T: 1 Total 10 letters.
There are 8 distinct types of letters: {A, S, I, G, N, M, E, T}.
The letters S and N are repeated twice.

Step 2: Consider cases for selecting 4 letters.
We need to choose 4 letters and then arrange them.
Case (a): All 4 letters chosen are distinct.
We have 8 distinct letters (A, S, I, G, N, M, E, T).
Number of ways to choose 4 distinct letters = \( {}^8C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \).
For each choice of 4 distinct letters, they can be arranged in \( 4! = 24 \) ways.
Number of words = \( 70 \times 24 = 1680 \).
Case (b): Two letters are alike, and the other two letters are distinct (and different from the alike pair).
There are 2 types of letters that can be chosen as the alike pair: S or N.
Subcase (b1): The alike pair is SS.
We need to choose 2 more distinct letters from the remaining 7 distinct letters (A, I, G, N, M, E, T).
Number of ways to choose these 2 letters = \( {}^7C_2 = \frac{7 \times 6}{2 \times 1} = 21 \).
The 4 letters are S, S, X, Y.
Arrangements = \( \frac{4!}{2!} = \frac{24}{2} = 12 \).
Number of words with SS = \( 21 \times 12 = 252 \).
Subcase (b2): The alike pair is NN.
Similarly, choose 2 more distinct letters from (A, S, I, G, M, E, T).
Number of ways = \( {}^7C_2 = 21 \).
The 4 letters are N, N, X, Y.
Arrangements = \( \frac{4!}{2!} = 12 \).
Number of words with NN = \( 21 \times 12 = 252 \).
Total for Case (b) = \( 252 + 252 = 504 \).
Case (c): Two letters are alike of one kind, and two letters are alike of another kind.
This means choosing the pair SS and the pair NN.
The 4 letters are S, S, N, N.
Number of arrangements = \( \frac{4!}{2!2!} = \frac{24}{4} = 6 \).

Step 3: Calculate the total number of such words.
Total number of words = Sum of words from all cases.
Total = (Words from Case a) + (Words from Case b) + (Words from Case c) Total = \( 1680 + 504 + 6 = 2190 \).
This matches option (4).