A random draw is being designed for 210 participants. A single winner is to be chosen, and all the participants must have an equal probability of winning. If the winner is to be drawn using 10 balls numbered 0 through 9, how many balls need to be picked, regardless of order, so that each of the 210 participants can be assigned a unique set of numbers?a) 10b) 4c) 5d) 3

Answer: The correct number of balls is (b) 4.Step-by-step explanation:  Given that a single winner is to be chosen in a random draw designed for 210 participants. Also, there is an equal probability of winning for each participant.We are using 10 balls, numbered through 0 to 9. We are to find the number of balls which needs to be picked up, regardless of order, so that each of the 210 participants can be assigned a unique set of numbers.Let 'r' represents the number of balls to be picked up.Since we are choosing from 10 balls, so we must have[tex]^{10}C_r=210.[/tex]The value of 'r' can be any one of 0, 1, 2, . . , 10.Now,if r = 1, then [tex]^{10}C_1=\dfrac{10!}{1!(10-1)!}=\dfrac{10!}{1!9!}=\dfrac{10\times 9!}{1\times 9!}=10<210.[/tex]If r = 2, then[tex]^{10}C_2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10\times 9\times 8!}{2\times 1\times 8!}=45<210.[/tex]If r = 3, then[tex]^{10}C_3=\dfrac{10!}{3!(10-3)!}=\dfrac{10!}{3!7!}=\dfrac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}=120<210.[/tex]If r = 4, then[tex]^{10}C_4=\dfrac{10!}{4!(10-4)!}=\dfrac{10!}{4!6!}=\dfrac{10\times 9\times 8\times\times 7\times 6!}{4\times 3\times 2\times 1\times 6!}=210.[/tex]Therefore, we need to pick 4 balls so that each participant can be assigned a unique set of numbers.Thus, (b) is the correct option.