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, a minimum of 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.) 6
C.) 5
D.) 4

Answer 1

Option D.

Explanation:

Total number of participants = 210

The probability of winning for each participant is equal.

We have 10 balls, numbered through 0 to 9 and we need 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 'x' represents the number of balls to be picked up. Total possible ways is defined by

`^(10)C_x`

where, x=0,1,2,3,4,5,6,7,8,9.

If x=10, then

`^(10)C_(10)=(10!)/((10-10)!10!)=1`

1 is less than 210.

If x=6, then

`^(10)C_(6)=(10!)/((10-6)!6!)=210`

Total 210 unique set of numbers.

If x=5, then

`^(10)C_(5)=(10!)/((10-5)!5!)=252`

252 is greater than 210.

If x=4, then

`^(10)C_(4)=(10!)/((10-4)!4!)=210`

252 is greater than 252.

It means 4 and 6 balls need to be picked so that each of the 210 participants can be assigned a unique set of numbers.

4 < 6

The minimum number of balls that need to be picked is 4.

Therefore, the correct option is D.

Answer 2

Option: D is the correct answer.

D) 4

Explanation:

A random draw is being designed for 210 participants.

If the winner is to be drawn using 10 balls numbered 0 through 9.

That is for the 210 participants we have to get a 210 unique numbers so that each participant winning chance is denoted by a unique number.This means that n balls are to be drawn out of the 10 balls such that we get total 210 choices irrespective of their order.

Hence, we have to find n such that:

`10_C_n=210`

A)

Now when n=10

we have:

`10_C_(10)=1\\eq 210`

Hence, option:A is incorrect.

B)

when n=6 we have:

`10_C_6=(10!)/(6!* (10-6)!)\\\\\\10_C_6=(10!)/(6!* 4!)\\\\\\10_C_6=210`

C)

n=5

`10_C_5\\\\=(10!)/(5!* (10-5)!)\\\\\\=(10!)/(5!* 5!)\\\\=252\\eq 210`

D)

n=4

`10_C_4=(10!)/(4!* (10-4)!)\\\\10_C_4=(10!)/(4!* 6!)\\\\10_C_4=210`

So, either 4 or 6 balls can be drawn in order to obtain 210 choices but we are asked to find the minimum number of balls and as 4<6 .

Hence, a minimum of 4 balls need to be drawn so that each receives a unique number.