A circular dartboard is labeled with each of the integers from 1 to 20

Question

A circular dartboard is labeled with each of the integers from 1 to 20 inclusive around its circumference. If the numbers are placed in random order around the circumference of the board, what is the probability that the number 1 is immediately next to the number 2?

A) \(\frac{1}{19!}\)

B) \(\frac{1}{20}\)

C) \(\frac{1}{19}\)

D) \(\frac{1}{10}\)

E) \(\frac{2}{19}\)

A)  \frac{1}{19!}

B)  \frac{1}{20}

C)  \frac{1}{19}

D)  \frac{1}{10}

E)  \frac{2}{19}

Abhishekgmat87 · Answer

Let us fix number 2. For number 1 to be next to number 2, it can be placed either to the left or to the right of number 2. There are 2 favourable ways to place number 1. Also probability of number 1 to be left of number 2 will be 1/19 as any 19 of the number can be left to number 2 and 1 is one such number. Similarly, the probability of number 1 to the right of number 2 will be 1/19.

Reqd probability = 1/19 + 1/19 = 2/19

Option E

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KarishmaB · Answer

What is the source of this question? I would worry about using the concept of circular arrangements on a dartboard or a clock. They have a natural point of reference on the wall which makes each position distinct. They can be put up in only one way so they have an up, down, left, right. 
When we talk about arrangement around a circular table, we are given that only arrangements which are different when considering relative placements to each other are valid i.e. the door, the window, the whiteboard etc are not points of reference. That is when we use the concepts of circular arrangement. In this question I would be tempted to mark the answer as 1/10.

mx39 · Answer

Karishma's comment aside (which is valid), here is how I approached this. I much prefer to use combinatorics on these kinds of questions; in fact, I'm not aware of any other method to solve for circular arrangements.

That said, recall that the probability here is always  \frac{# of favorable outcomes}{# of total outcomes} . Let's tackle the first—the total number of outcomes—to get it out of the way.

Total Number of Outcomes 
With circular arrangements, the formula is very simple: the total number of arrangements is  (n - 1)! , where  n  = the number of entities. With 20 distinct numbers, the total number of arrangements is  19! .

Number of Favorable Outcomes 
Now, the slightly trickier bit. There are two things to keep in mind here: first, that we are completely agnostic as to whether 2 comes before 1, or vice-versa. Second, that this is still a circular arrangement.

Recall the formula for calculating the number of arrangements when two items must be together:  (n)!(y - n + 1)! ; where  n  is the number of elements we are keeping together and  y  is the total number of elements. In this case, it would be  2!(20-2+1)! , or  2!19! .  However, this is a circular arrangement, meaning we must subtract 1 from the total number of arrangements, for 2!18!.

Combine into a Fraction 
Glancing at the answer choices, we're clearly looking for a simplified fraction here, so the name of the game is expressing the unsimplified fraction and paring that down. Given that probability =  \frac{# of favorable outcomes}{# of total outcomes} , our fraction is  \frac{2!18!}{19!} . Our answer choices are denominated either in terms of a 19 or a 20, so let's try and get there (this means AC A is completely out since there's no way our fraction will ever get to be 1/19!).

If we rewrite our fraction as  \frac{2!18!}{19 * 18!} , something becomes clear immediately. The 18! will cancel out to become  \frac{2!}{19} , or simply  \frac{2}{19} ,   leaving AC E as the correct answer   .

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A circular dartboard is labeled with each of the integers from 1 to 20

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A circular dartboard is labeled with each of the integers from 1 to 20

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