If 6n tickets numbered 0,1,2,...,6n-1 are placed in a bag, and three are drawn out , show that the chance that the sum of the numbers on them is equal to 6n is
\frac{3n}{(6n-1)(6n-2)}
Looking for simpler and better ways
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3 Answers
30Consider the equation,
x+y+z=6n where x,y,z ≥ 0
No. of solutions = 6n+2C2
Now, these solutions include these as well (0,0,6n), (0,6n,0) and (6n,0,0).
Eliminating these three cases and eliminating the ordering, probability= \frac{^{6n+2}C_{2}-3}{3!}\times \frac{1}{^{6n}C_{3}}=\frac{^{6n+2}C_{2}-3}{6n(6n-1)(6n-2)}
which comes down to something like \frac{(3n+1)(6n+1)-3}{6n(6n-1)(6n-2)}
Could someone point out where I am going wrong? [7]
21x+y+z= 6n
number of solution that you have got includes cases where all the variables are equal, any two are equal and their permutations.
as three tickets are drawn clearly those cases can not hold
i subtracted those cases and managed to get the answer
thatz y i m looking for a simpler and better solution..