A) total number of integers which are less than 1000 and have distinct digits
= number of 3 digit numbers which have distinct digits + number of 2 digit numbers which have distinct digits + number of 1 digit numbers which have distinct digits
= (9*9*8) + (9*9) + 9
= 648 + 81 + 9
= 738
B) total number of integers which are less than 1000 and have distinct digits and are even
= number of 3 digit numbers which have distinct digits and are even + number of 2 digit numbers which have distinct digits and are even + number of 1 digit numbers which have distinct digits and are even
= [ (9*8*1) + (8*8*4) ]+ [ (9*1) +(8*4) ] + [4]
= 72 + 256 + 9 + 32 + 4
= 373
NOTE : IN 2 ND QUESTION , I HAVE DIVIDED 3 DIGIT NUMBERS INTO 2 CASES
1) LAST DIGIT IS 0 (HIGHLIGHTED WITH BLUE)
2) LAST DIGIT IS NOT 0
SIMILIARLY FOR 2 DIGIT NUMBERS ALSO I HAVE DIVIDED INTO TWO CASES.