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Test of Mathematics Solution Subjective 56 - Number of Four Digit Integers

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 56 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

 Show that the number of ways in which four distinct integers can be chosen from 1,2,...n $ (n \ge 7) $ such that no two are consecutive is equal to ${{n-3}\choose{4}}$.

Solution:

suppose we chose a>b>c>d four integers from 1,2,...n such that no two are consecutive.So d<c-1<b-2<a-3.

Now d,(c-1),(b-2),(a-3) are four distinct integers from 1,2,...(n-3),which may be consecutive or not.

So number of ways to chose four numbers such that no two are consecutive from 1,2,...n = number of ways to chose four distinct numbers from 1,2,...(n-3).

Now number of ways to chose 4 ways from 1,2,...(n-3) =

$ {{n-3}\choose{4}} $.

Conclusion: Number of ways to chose four distinct integers from 1,2,...n such that no two are consecutive is $ {{n-3}\choose{4}} $

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 56 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

 Show that the number of ways in which four distinct integers can be chosen from 1,2,...n $ (n \ge 7) $ such that no two are consecutive is equal to ${{n-3}\choose{4}}$.

Solution:

suppose we chose a>b>c>d four integers from 1,2,...n such that no two are consecutive.So d<c-1<b-2<a-3.

Now d,(c-1),(b-2),(a-3) are four distinct integers from 1,2,...(n-3),which may be consecutive or not.

So number of ways to chose four numbers such that no two are consecutive from 1,2,...n = number of ways to chose four distinct numbers from 1,2,...(n-3).

Now number of ways to chose 4 ways from 1,2,...(n-3) =

$ {{n-3}\choose{4}} $.

Conclusion: Number of ways to chose four distinct integers from 1,2,...n such that no two are consecutive is $ {{n-3}\choose{4}} $

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