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

• 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
• 9 years agoHelpfull: Yes(64) No(8)
• 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!
• 9 years agoHelpfull: Yes(4) No(6)
• 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.
• 9 years agoHelpfull: Yes(3) No(11)
• 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
  
• 9 years agoHelpfull: Yes(2) No(5)
• 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.
• 9 years agoHelpfull: Yes(1) No(8)
• 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.
  
• 9 years agoHelpfull: Yes(1) No(4)
• 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
• 9 years agoHelpfull: Yes(1) No(5)