IIT JEE , AIEEE Forum - Mathematics , Algebra

magiclko

n1 and n2 are two 5-digit numbers. Total no of ways of forming n1 and n2, so that these numbers can be added without carrying at any stage, is equal to

a) 36 (55)^4

b) 45 (55)^4

c) (55)^5

d) none of these

kusum

hi ,
For each digit of the largest no. of the 2 we have exactly the same no. of digits that can b filled in that place of the 2nd no.
So total possible combinations=1+2+3+.........+10=55
So ways of choosing any one out of these 55 combinations=C(55,1)=55
This is valid for all the 8 digits(except the 1st) but not for the 1st digit as it can't take 0 value.
So possible combinations for this=1+2+.....+9=45
So the required no. of ways =45*(55)^8

rik_mad

36(55)^4 shud be the answer

this is because 0 will not be a part of both the numbers beginning (left most) position... so number of possibilities reduces from 55 to 36

KAB

I think Rik mad is correct.

magiclko

yeah rik_mad is right!!!!... thanx

amar.gupta

Good work rik_mad

Ask & Discuss Questions with Community & Experts