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