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In how many ways, two different natural numbers can be selected, whichless than or equal to 100 and differ by almost 10. |
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Answer» Correct Answer - `.^(100)C_(2)-.^(90)C_(2)` 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)`. |
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