Question 397256: Consider the numbers between 100 and 400. How many of these numbers are even and have no repetition of digits?
Case 1: 2 or 3 in first position and 0 in last position: 2x8x1 = 16
Case 2: 1 or 3 in first position and 2 in last position: 2x8x1 = 16
Case 3: 1,2 or 3 in first position and 4,6,8 in last position: 3x8x3 = 72
Total:104....BUT the answer is 112. What am I missing?
Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website! case 1: when 1 at first place
no. of ways to put even digit at third position = 5
no. of ways to put any remaining digit at second position = 8
(except 1 and digit at third position)
total no. of possible numbers = 5*8 = 40
case 2: when 3 at first place, similar to previous condition
total no. of possible numbers = 5*8 = 40
case 3: when 2 at first place,
no. of ways to put even digit at third position = 4 (except 2)
no. of ways to put any remaining digit at second position = 8
(except 1 and digit at third position)
total no. of possible numbers = 4*8 = 32
total = 40+40+32 = 112
try to understand the concept .........
for any query you are welcome to contact by e-mail..
Mistake in your solution...
in case 1 (when 0 at last place)
first position can be filled by 3 ways (1,2 or 3).
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