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Out of 21 tickets marked with numbers from 1 to 21, three are drawn at random. What is the probability that the numbers on them will be in AP?
let the numbers be a,b,c so if a,b,c are in AP the condition is
b=a+c/2
it means that b is an integer ,so from here we can if a,b are even or a,b are odd because odd+odd=even or even+even=even
so consider case 1 where both a and b are odd numbers so the possible numbers of outcomes is 11C2 because there are 11 odd numbers between 1 to 21
now consider case no.2 where a and b are even numbers so total possible ways is 10C2 because there are 10 even numbers between 1 to 21
if we get a and b then we can easily calculate the b so we dont need to calculate the total ways of selecting b
and the total no. of outcomes are 21C3
so probibality =[10C2 + 11C2]/21C3=10/133
ans -- 10/119 , maximum common difference can be 10 (tickes nos selected (1 , 11 , 21) ) and minimum common difference is 1 , comm difference ---- count of triplets with that common difference 1
common diff ---- count of triplets with that common diff
1 19
2 17
.......
10 1
this will form an AP
TOTAL = 1+3.+5+7...........17+19 = 100
possible ways -- 21C3
ans -- 10/119
