CBSE Important Questions

Mathmatics Class IX

is  total possibility in the under (ii) is

4+3+2+1
+3+2+1
+2+1
+1
20

Similarly for unit digit 6

The number of numbers are

with starting digit
1 = 6+5+4+3+2+1
2 = 5+4+3+2+1
3 = 4+3+2+1
4 = 3+2+1
5 =2+1
6 =1

Total = 56

Similarly for unit digit 8

The number of numbers are

With starting digit
1 = 8+7+6+5+4+3+2+1
2 = 7+6+5+4+3+2+1
3 = 6+5+4+3+2+1
4 = 5+4+3+2+1
5 = 4+3+2+1
6 = 3+2+1
7 = 2+1
8 = 1

Total = 120

So total possible number = 4+20+56+120 = 200 Ans

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