When a + b = 1, possible values of a , b is (1, 0) and c, d is (1,0) , (0, 1)
Total possible numbers abcd = 1 x 2 = 2 (i.e 1001, 1010)
(Remember the concept if part of a work can be done first by m diff ways and then by n diff ways then the total number of ways the work can be done is mxn)
When a + b = 2, possible values of a , b is (2, 0), (1,1) and c, d is (2,0) , (0, 2), (1,1)
Total possible numbers abcd = 2 x 3 = 6 (i.e 2002, 2020, 2011, 1120, 1102, 1111)
When a + b = 3, possible values of a , b is (3,0), (2, 1), (1,2) and c, d is (3,0), (0, 3), (2, 1), (1,2)
Total possible numbers abcd = 3 x 4 = 12 (i.e 2002, 2020, 2011, 1120, 1102, 1111)
So its like for any one digit number n, as sum of (a + b) there‘ll be n different values of (a, b) : (n, 0), (n-1), (n-2). . . . (1,n) and corresponding values of (c , d) is n + 1 : (n, 0), (n-1), (n-2). . . . (1,n), (0, n)
So we get for a + b <= 9 possible values =
1x2 + 2x3 + 3x4 + 4x5 + . . . . . . . . + 9x10
= ∑[n x (n+1)] for n=1 to 9
= ∑n2 + ∑ n for n=1 to 9
= 1/6 * 9 * 19* 10 + 45
= 285 + 45
=330
Now lets consider for a + b >= 10
For a + b = 10, possible values are (1+9, 2+8, 3+7, 4+6, . . . . 9 + 1)
Number of possible values = 9 and corresponding number of values of c, d is 9 : (1+9, 2+8, 3+7, 4+6, . . . . 9 + 1)
Total possible numbers abcd with sum 10 = 9 x 9
Similarly total possible numbers abcd with sum 11 = 8 x 8 (2+9, 3+8. . . .9+2)
. . . . . . .
Similarly total possible numbers abcd with sum 18 = 1 x 1 (9+9)
Number of possible values = 9 and corresponding number of values of c, d is 9 : (1+9, 2+8, 3+7, 4+6, . . . . 9 + 1)
Total possible numbers abcd with sum 10 = 9 x 9
Total possible values of abcd for a +b >=10 is
Hope I have not done any calculation mistake.
n/a
@ Nishit
Perfect approach.................
Thanks!
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