There is a five digit code consist of letters and digit. the code can have one digit from 1,2,3 and any four letters from A,b,c,d,e. first and last code must be a letter and all letters can be repeated. How many combinations can be possible?

A. 625
b. 3840
c. 5625
d. 11,520

• since repetitions are allowed so there are 5 ways to choose the letters for the first slot, and five ways to choose for last slot
  
  that can be like:- 5 x _ x _ x _ x 5
  
  you still have to choose 1 digit and 2 letters between. For the letters, again, repetitions are allowed so each slot can be filled in 5 ways. For digit, it can only be filled in 3 ways. However, since the digit can be in any one of the 3 middle slots, so we multiply by 3 again:
  
  5 x 3 x 5 x 5 x 5 or
  5 x 5 x 3 x 5 x 5 or
  5 x 5 x 5 x 3 x 5
  
  so 5 x 5 x 5 x 5 x 3 x 3 = 5^4 x 3^2 = 5625.
• 9 years agoHelpfull: Yes(32) No(1)
• 5c1*(5c1*5c1*3c1)*3!/2!*5c1=5625
  
  First fix the 1st and last digit, that can be filled with any of the 5 letters in 5c1 ways
  Remaining 3 digits includes a no which can be filled in 3c1 ways
  Remaining 2 digits are again letters and can be filled in 5c1 ways
  these 3 digits can be arranged among themselves in 3!/2! ways
  5625 is the answer!
• 9 years agoHelpfull: Yes(21) No(4)
• ans will be 11,520
  Choose one number from 3 numbers in 3C1 = 3 ways
  Choose four letters from 5 letters in 5C4 = 5 ways
  Choose one position from the middle three for the number in 3C1 = 3 ways
  The other four positions can be filled by the 4 letters in 4^4 ways.
  
  Therefore total number of codes possible = 3*5*3*(4^4) = 45*16*16 = 11,520
• 9 years agoHelpfull: Yes(5) No(9)
• Choose one number from 3 numbers in 3C1 = 3 ways
  Choose one position from the middle three for the number in 3C1 = 3 ways
  The other four positions can be filled by the 5 letters in 5^4 ways.
  
  Therefore total number of codes possible = 3*3*(5^4) = 5,625
• 9 years agoHelpfull: Yes(1) No(0)
• 5c1*(5c1*5c1*3c1)*3!/2!*5c1=5625
  
  First fix the 1st and last digit, that can be filled with any of the 5 letters in 5c1 ways
  Remaining 3 digits includes a no which can be filled in 3c1 ways
  Remaining 2 digits are again letters and can be filled in 5c1 ways
  these 3 digits can be arranged among themselves in 3!/2! ways
  5625 is the answer
• 5 years agoHelpfull: Yes(0) No(0)