Number of 3-digit numbers in which digit at hundred's place is greater
than other two digits.
MY ATTEMPT:
Two cases are possible:
1) Ten's and Unit's digits are identical
$9+8+7+…1=45$ possibilities
2) Ten's and Unit's digits are not identical
$\frac{1}{3}{(9.9.8)}$ possibilities $=216$
Total $45+216=261$ but the real answer is $285$.Where did I go wrong?
Best Answer
It doesn't matter if the tens and unit digits are different. Just that they are less than the hundreds digit. So for 9 any number 0-8 in tens or one digit is acceptable. 9X9 is 81. For 8 any number between 0-7 is acceptable. 8x8 is 64. working way down to 1 only 0 in both tens and ones works. That will sum to 285.