In how many ways, two different natural numbers can be selected, which less than or equal to 100 and differ by almost 10.

Answer

`.^(100)C_(2)-.^(90)C_(2)`

Explanation

Number of ways in which two numbers can be selected from 100 integers is `.^(100)C_(2)`.
Let us find the number of ways by selecting two numbers, if their difference is more than 10.
Let the two integers selected be P and Q.
P Q
x y z
Here, x,y,z are number of integers before P, between P and Q, and after Q respectively. Therefore, x+y+z=98.
Now, we have to find number of integral solution for `x,z ge 0` and `y ge 10`.
`therefore x+y_(1)+z=88, " where" y_(1)ge 0`
Then, number of solutions of above equations is
`.^(88+3-1)C_(3-1)= .^(90)C_(2)`.
Hence, the number of ways where two different natural numbers can be selected, which differ by almost 10 is `. ^(100)C_(2)- .^(90)C_(2)`.