24Sir I have got these conditions..but not able to get hte final answer match with boook...tell me what ur answer is ???
Q1 Let n and m be any two 3 digit numbers .Find total number of pairs of n and m so that n and m can be added without carrying
Q2Let n and m be any two 3 digit numbers .Find total number of pairs of n and m so that n can be subtracted from m without borrowing
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9 Answers
62The first question seems to be simple...
abc, xyz
c+z<10, a+x<10, b+y<10
given at a>0, x>0
b, c, y, z>=0
Now you can solve?
2462c+z<10, a+x<10, b+y<10
given at a>0, x>0
b, c, y, z>=0
a=1, x=1, 2, ... 8 .... 8 ways
a=2 7 waysl...
a=8.... 1 way
total a, x can be selected in 36 ways......
b=0, x=0....9 10 ways...
b=1, x=0,1,..8 9 ways
total b, y in 55 ways
c and z in 55 ways
total ways = 36*55*55 ways (given that you assume that 123+234 is different from 234+123
otherwise you have to eliminate the cases when both abc and xyz are equal
1if in first case the the similar cases are not to be included then the ans will becomes half (as every case is repeated twice)
62No Gaurav sir.. it is only for the first part...
The second part will be slightly different..
btw for the second part, a simpler method will be that
chose any two digits, there will be one way to chose x and a, because we want a<x, b<y and c<z
so we have to chose 2 digits from 1,2.... 9 for first digit = 9x9 = 81
2nd digit can be chosen from 0,1,2....9 = 10x10
3rd digit can be chosen from 0,1,2....9 = 10x10
total ways 8,10,000