How many different 4-letter words can be made using the letters of the

Question

How many different 4-letter words can be made using the letters of the word TENNESSEE ?

A. 47
B. 139
C. 163
D. 171
E. 3780

gmatophobia · Answer

T 
N N
S S
E E E E

We require four letter words. To do so, we need to follow two step process

Step 1: Select four alphabets
Step 2: Arrange the selected alphabets.

Case 1:   No repetition

Step 1 : There are four distinct alphabets. The alphabets can we selected in 4C4 ways = 1.

Step 2 : The distinct alphabets can be arranged in 4! ways.

Total : 4! = 24

Case 2:   One repeating pair

Step 1 : The repeating pair can be selected from three available alphabets (E, N and S) .

As we need only one alphabet from the available three repeating alphabets, the selection can be done in 3C1 = 3 ways

Once we have have selected the repeating pair, we need two distinct alphabets from the three available 3 alphabets.

This selection can be done in 3C2 = 3 ways.

Step 2 :

One repeating pair and two distinct alphabets can be arranged in  \frac{4! }{ 2!}  ways

Total :  3 * 3 * \frac{4! }{ 2! }  = 108 ways

Case 3:   Two repeating Pairs

Step 1 : Two alphabets that repeat can be selected from three available alphabets (E, N and S) .

We need only two alphabet from the available three repeating alphabets, the selection can be done in 3C2 = 3 ways

Step 2 :

Two repeating pair can be arranged in  \frac{4! }{ 2! * 2! }  ways

Total :  3 *\frac{4! }{ 2! * 2! }  = 18 ways

Case 4:   Three Alphabets repeat

Step 1 : The only possible alphabet that can repeat thrice is E.

We still need the fourth distinct alphabet, which can be selected from the three available alphabets in 3C1 ways.

Step 2 :

The repeating pair and one distinct alphabet can be arranged in  \frac{4! }{ 3! }  ways

Total :  3 *\frac{4! }{ 3! }  = 12 ways

Case 5:   Four Alphabets repeat

Step 1 : The only possible alphabet that can repeat four times is E.

Step 2 :

Number of ways to arrange four "E"s= 4!/4! = 1

Total : 1 way

Sum = 24 + 108 + 12 + 18 + 1= 163

Option C

Ipsita! · Answer

Can someone explain why we are multiplying 3 everytime?

Posted from my mobile device

chetan2u · Answer

Hi

It is not being multiplied by 3 every time, but with the ways certain digits can be taken.
1) All same digit - Only one E, as no other digit is available.
2) Two digits: 
Case A: Both digits are taken two times. X,X,Y,Y.....There are 3 digits that are available, E, N and S. Choosing two out of them for X and Y will mean 3C2.
Case B: 1 is used 3 times and other one time. X,Y,Y,Y....E can be used 3 times, so Y=E, and the remaining spot, X, can be filled by any of the remaining 3, T, N and S. so 3C1

Similarly for others too.

WiziusCareers1 · Answer

To solve this problem, we first need to look at the frequency of each letter in the word TENNESSEE:
E: 4
N: 2
S: 2
T: 1
Total letters: 9 (with 4 distinct types: E, N, S, T)

We are choosing 4 letters to form a "word" (permutation).

Because we have multiple repetitions, we must break the problem down into cases based on the combinations of letters chosen.

Case 1: 4 of a kind (All 4 letters are the same) 
The only letter available 4 times is E.
Combination: {E, E, E, E} → 1 way
Permutations: {4!}/{4!} = 1 way

Case 2: 3 of a kind + 1 different 
We must pick the letter that appears 3+ times (only E) and then 1 letter from the remaining 3 types {N, S, T}.
Combinations: 3C1*1C1 = 3 ways ({E,E,E,N}, {E,E,E,S}, {E,E,E,T})
Permutations: 3 * 4!/(3!*1!) = 12 ways

Case 3: 2 of a kind + 2 of a kind (Two pairs) 
We need to choose 2 types of letters from those that appear at least twice {E, N, S}.
Combinations: 3C2 = 3 ways ({E,E,N,N}, {E,E,S,S}, {N,N,S,S})
Permutations: 3 * 4!/(2! * 2!) = 18 ways

Case 4: 2 of a kind + 2 different 
First, pick the letter for the pair from {E, N, S}. Then, pick 2 different letters from the remaining 3 types.
Combinations: 3C1 * 3C2 = 9 ways
Permutations: 9 * 4!/(2!*1!*1!) = 108 ways

Case 5: All 4 different 
We must pick one of each of the 4 distinct letters {E, N, S, T}.
Combinations: 4C4 = 1 way {E, N, S, T}
Permutations: 1 * 4! = 24 ways

Total 
Adding the permutations from all possible cases: 1 + 12 + 18 + 108 + 24 = 163
 The correct answer is C. 163

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