Question

How many 4-letter permutations can you form from the letters of equations and conjugates?

How many distinct permutations can be made from the letters of each word?
statistics , independence and combinatorics?

Answer 1

Answer:

2454

Step-by-step explanation:

The word 'EXAMINATION' contains 2 A’s, 2 I’s, 2 N’s, and 5 other letters.

We can break down the number of 4 letter words that can be formed into the following categories:

Words having no double letters.

We have four places to fill in a four letter word _ _ _ _.

If we do not want double letters in the word, then we have 8 different letters (E, X, A, M, I, T, O, N) to choose from.

Thus, number of arrangements of 8 letters, taking 4 at a time = 8P4 = 1680

Words having one double letter.

One of the three double letters (A, I, N) can be chosen in 3C1 ways.

Suppose if letter ‘A’ is chosen and we position the A’s at any two spots (_ _ A A), then we have only two blank spaces left. Out of the remaining 7 letters (E, X, M, I, T, O, N), two places can be filled in 7P2 ways. Also, the two A’s can be arranged at the 4 spots in 4C2 ways.

Number of arrangements for letter A = 7P2. 4C2

Thus, total number of arrangements = 7P2. 4C2×3 = 756

Words having two double letters.

Two of the three double letters can be chosen in 3C2 ways. Also, the two chosen letters can be arranged at the four places in 4C2. 2C2 ways.

Thus, number of arrangements = 3C2 × 4C2. 2C2 = 18

Hence, total number of permutations = 1680+756+18 = 2454