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(asking for the second time!(Permutation and Combination We have integers from 1 to 100 Number of ways of selecting two numbers such that there difference is atmost 10. I did this by bijection method and I am getting the answer 100C2-91C2. I don’t know the answer as “None of these” is the correct answer But I have to confirm. So please someone solve it and tell me the answer.

(asking for the second time!(Permutation and Combination
We have integers from 1 to 100
Number of ways of selecting two numbers such that there difference is atmost  10.
I did this by bijection method and I am getting the answer 100C2-91C2.
I don’t know the answer as “None of these” is the correct answer But I have to confirm.
So please someone solve it and tell me the answer.

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1 Answers

Vijay Mukati
askIITians Faculty 2590 Points
7 years ago
Dear Student, If you choose any number between1and90(inclusive), and call itx, you can choose any number betweenx+1andx+10, which gives us10choices for each. If you choose91, for example, you have9 numbers to choose from (i.e.92,93,⋯100). Similarly,93gives us8choices,94gives us7, and so forth, until99gives us only1choice. Hence,90x10+9+8+7+⋯+1=945.

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