How many distinct 4-letter “words” (with or without meaning) can be ma

Question

How many distinct 4-letter “words” (with or without meaning) can be made by using the eleven letter word “EXAMINATION”?

A. 7290

B. 2454

C. 1680

D. 912

E. 330

Kalpit1212 · Answer

We have 8 different digits, 3 repeated digits

1) When all digits are different : 8P4 = 1680
2) When one digit is repeated: 3C1 x 7C2 x 4!/2! = 756
3) When 2 digits are repeated : 3C2 x 4!/2!2! = 18

Hence, total = 2454

olaflesniak1998 · Answer

Could you maybe elaborate why you in case 2 and 3 multiply by factorial 4! ?

gmatophobia · Answer

We have the following combinations of alphabets available with us -

E X M T O A A I I N N

Question : Create four-letter words with or without meaning.

The creation of a letter involves the following two steps -

Step 1 : Select four letters

Step 2 : Arrange those four letters

Case 1  : No letters repeat (4 distinct alphabets)

Step 1 : Select four letters

There are 8 distinct letters, hence we can select 4 letters in  ^8C_4  ways.

Step 2 : Arrange those four letters

The 4 distinct letters so selected in step 1, can be arranged in 4! ways.

Possible words :  ^8C_4  * 4! = 1680

Case 2  : One pair of alphabet repeats (2 distinct alphabets & 1 repeating pair)

Step 1 : Select four letters

The repeating pair can be selected in  ^3C_1  ways

From the remaining 7 distinct alphabets, we can select two alphabets in  ^7C_2  ways

Step 2 : Arrange those four letters

The word is of the form XXYZ, and the number of possible arrangements =  \frac{4!}{2!}

XX is the repeating pair
Y & Z are unique alphabets

Possible words :  ^3C_1 * ^7C_2 * \frac{4!}{2!}  = 756

Case 3  : Two pairs of alphabet repeats (2 pairs of repeating alphabets)

Step 1 : Select four letters

The repeating pairs can be selected in  ^3C_2  ways

Step 2 : Arrange those four letters

The word is of the form XXYY, and the number of possible arrangements =  \frac{4!}{2!*2!}

XX & YY are the repeating pairs

Possible words :  ^3C_2 * \frac{4!}{2!*2!}  = 18

Total of all three cases = 1680 + 756 + 18 = 2454

Option B

Hope this clarifies your question!

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How many distinct 4-letter “words” (with or without meaning) can be ma

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