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

Answer:

Option D.

Step-by-step 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

$^{10}C_{10}=\dfrac{10!}{(10-10)!10!}=1$

1 is less than 210.

If x=6, then

$^{10}C_{6}=\dfrac{10!}{(10-6)!6!}=210$

Total 210 unique set of numbers.

If x=5, then

$^{10}C_{5}=\dfrac{10!}{(10-5)!5!}=252$

252 is greater than 210.

If x=4, then

$^{10}C_{4}=\dfrac{10!}{(10-4)!4!}=210$

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

Answer:

Option: D is the correct answer.

D) 4

Step-by-step 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:

$10_C_{10}=1\neq 210$

Hence, option:A is incorrect.

B)

when n=6 we have:

$10_C_6=\dfrac{10!}{6!\times (10-6)!}\\\\\\10_C_6=\dfrac{10!}{6!\times 4!}\\\\\\10_C_6=210$

C)

n=5

$10_C_5\\\\=\dfrac{10!}{5!\times (10-5)!}\\\\\\=\dfrac{10!}{5!\times 5!}\\\\=252\neq 210$

D)

n=4

$10_C_4=\dfrac{10!}{4!\times (10-4)!}\\\\10_C_4=\dfrac{10!}{4!\times 6!}\\\\10_C_4=210$

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.