Question

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) 10
b) 4
c) 5
d) 3

Answer 1

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

$^{10}C_r=210.$

The value of 'r' can be any one of 0, 1, 2, . . , 10.

Now,

if r = 1, then

$^{10}C_1=\dfrac{10!}{1!(10-1)!}=\dfrac{10!}{1!9!}=\dfrac{10\times 9!}{1\times 9!}=10$

If r = 2, then

$^{10}C_2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10\times 9\times 8!}{2\times 1\times 8!}=45$

If r = 3, then

$^{10}C_3=\dfrac{10!}{3!(10-3)!}=\dfrac{10!}{3!7!}=\dfrac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}=120$

If r = 4, then

$^{10}C_4=\dfrac{10!}{4!(10-4)!}=\dfrac{10!}{4!6!}=\dfrac{10\times 9\times 8\times\times 7\times 6!}{4\times 3\times 2\times 1\times 6!}=210.$

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.

Answer 2

Answer:

Option b - 4

Step-by-step explanation:

Given : 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.

To find : 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?

Solution :

Let n be the number of balls drawn.

According to the question,

n balls are to be drawn out of the 10 balls such that we get total 210 choices irrespective of their order i.e. $^{10}C_n=210$

Now we check fro given options,

a) The value of n=10

$^{10}C_{10}=\frac{10!}{10!\times 0!}$

$^{10}C_{10}=1\neq 210$

It is not correct.

b) The value of n=4

$^{10}C_{4}=\frac{10!}{4!\times (10-4)!}$

$^{10}C_{4}=\frac{10\times 9\times 8\times 7\times 6!}{4\times 3\times 2\times 6!}$

$^{10}C_{4}=210$

It is correct.

c) The value of n=5

$^{10}C_{5}=\frac{10!}{5!\times (10-5)!}$

$^{10}C_{5}=\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times4\times 3\times 2\times 5!}$

$^{10}C_{4}=252\neq 210$

It is not correct.

d) The value of n=3

$^{10}C_{3}=\frac{10!}{3!\times (10-3)!}$

$^{10}C_{3}=\frac{10\times 9\times 8\times 7!}{3\times 2\times 7!}$

$^{10}C_{4}=120\neq 210$

It is not correct.

Therefore, option b is correct.