Permutation and combination

Asked by nidhilmohan | 27th Feb, 2010, 07:30: PM

Expert Answer:

Y1Y2Y3 - X1X2X3, and so that we need no borrowing.

This sets a condition on the digits X1, X2, and X3 such that,

X1Y1,  X2Y2,  X3Y3.

So X1 can be selected in 0, 1, 2, ...., Y1 ways.

So X2 can be selected in 0, 1, 2, ...., Y2 ways.

So X3 can be selected in  1, 2, ...., Y3 ways.

Hence the total number of such  X1X2X3 numbers is, (Y1+1)(Y2+1)Y3.

Also this is also the total number of ways of doing the described subtraction.





Answered by  | 28th Feb, 2010, 12:23: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.