Question
Thu April 17, 2014 By: Viresh Mohan

 

Expert Reply
Sat April 19, 2014
 
 
We can choose 3 tickets out of 21 tickets in C(21,3) ways.
 
The three-term A.P. is determined by its first and last elements, which must have the same parity.
 
 
There are 10 even numbers and 11 odd numbers out of 21.
 
So, if we count the ways to pick the highest and the lowest such that they have the same parity, then, the middle number can be selected in a specific way.
 
If the lowest is 1, then the middle number can be selected in 10 ways.
 
If the lowest is 2 or 3, then the middle number can be selected in 9 ways.
 
If the lowest is 4 or 5, then the middle number can be selected in 8 ways.
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If the lowest is 10, then the middle number can be selected in 1 ways.
 
So the total
=n(n+1)2+(n1)n2=n2 
=n(n+1)2+(n1)n2=n2 
 
equals 10 plus 2 cross times 9 plus 2 cross times 8 plus 2 cross times 7 plus... plus 2 equals 10 plus 9 plus 8 plus 7 plus... plus 1 plus 9 plus 8 plus 7 plus... plus 1 equals fraction numerator 10 open parentheses 11 close parentheses over denominator 2 end fraction plus fraction numerator 9 open parentheses 10 close parentheses over denominator 2 end fraction equals 55 plus 45 equals 100
 
Total number of arrangements = C(21,3)=1330
 
So, the probability of the tickets that are drawn in AP
equals 100 over 1330 equals 10 over 133
=n(n+1)2 +(n1)n2 =n2
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