Latest Elitmus Aptitude Question SOLUTION: How many 6 letter word can be formed by the word mediterranean such that start with E and end with R?

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(#M40151795) ELITMUS Numerical Ability question P&C Keep an EYE

How many 6 letter word can be formed by the word mediterranean such that start with E and end with R?

Asked In Elitmus NRJ Soni (9 years ago)

Solved lokesh Read Solution (7)

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Read Solution (7) :

lokesh  (9 years ago) Best Solution

2454

The first letter is E and the last one is R.

Therefore, one has to find four more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters(M , D , I ,T ,R).

The four positions can either have four different letters or 1 pair of same letters or have 2 pairs of same letters.

Case 1: When the FOUR letters are different. One has to choose FOUR different letters from the 8 available different choices. This can be done in 8 * 7 * 6 * 5= 1680 ways.

Case 2: When the we have one pair of same letter and 2 different letter.
=> selecting 1 pair out of 3 and two letters from the remaining 7 different options
=> 3 * 7C2
now possible arrangements of these four letters in which 2 are identical is 4! / 2! = 12

which gives total ways = 12 * 3 * 21 = 756
Case 3: When 2 pairs of same letters are selected
selecting two pairs out of three = 3 ways
now possible arrangements = 4! / 2! * 2! = 6

which gives = 6 * 3 = 18 ways

So, Total number of posssibilities = 1680 + 756 + 18 = 2454

Is this solution Helpfull? Yes (64) | No (8)  

annu  (9 years ago)

Total number =13
Word starts with e and end with r hence 11 letter left
Here e , a , n are 2 times(in remaining 11 letters) hence solution is 11p4 /2!*2!*2!

Is this solution Helpfull? Yes (4) | No (6)  

saambuddha bag  (9 years ago)

total no of letter=13
starting with E and end with R of a 6 letter word means, there r 4 place in between start and end.we have already taken 1 "E" and 1 "R". rest 11 letter can be placed like 11p4. that is 7920.

Is this solution Helpfull? Yes (3) | No (11)  

monika talwar  (8 years ago)

after fixing E and R at 1st and 6th position we left with 4 letters....If the first letter is M, D, I, T or R, we have 7 choices for the second letter,that's 5*7 = 35 words.
if the first letter is (E A 0r N) then we have 3*8=24 words
so ans is 59

Is this solution Helpfull? Yes (2) | No (5)  

Souparno Mukherjee  (9 years ago)

ans. 78
there are 2 N's , 2A's, 2E's, R, M, D & I left for the rest 4 places.
now 4 places can be filled in
case 1: taking 2 pairs - 4! / (2!* 2!) and can be mutually done in 3 ways so 3(4!/(2!*2!)).
case 2: 1 pair and 2 different- 4!/2!.
case 3: 4 different - 4!.
total no. of arrangements 78.

Is this solution Helpfull? Yes (1) | No (8)  

Souparno Mukherjee  (9 years ago)

ans. 78
there are 2 N's , 2A's, 2E's, R, M, D & I left for the rest 4 places.
now 4 places can be filled in
case 1: taking 2 pairs - 4! / (2!* 2!) and can be mutually done in 3 ways so 3(4!/(2!*2!)).
case 2: 1 pair and 2 different-3( 4!/2!).
case 3: 4 different - 4!.
total no. of arrangements 78.

Is this solution Helpfull? Yes (1) | No (4)  

akash  (8 years ago)

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of posssibilities = 56 + 3 = 59

Is this solution Helpfull? Yes (1) | No (5)  

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