If there are 7 doors in a room how many ways can a person enter one door and leave by another door

The GRE maths sample question given below is a problem solving question in permutation combination. This GRE sample question is an easy question.

Question 1: There are 5 doors to a lecture room. In how many ways can a student enter the room through a door and leave the room by a different door?

  1. 10
  2. 9
  3. 20
  4. 625
  5. 1024

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Explanatory Answer

Step 1 of solving this GRE Permutation Question: Find the number of ways the student can enter the classroom

There are 5 doors to the classroom. The student can enter the class through any one of these doors.
It could be any one of A or B or C or D or E. So, there are 5 ways.

Step 2 of solving this GRE Permutation Question: Find the number of ways the student can leave the classroom through a different door.

The student cannot leave the classroom through the door she entered.
The number of choices to choose a door to leave is down to 4

Step 3 of solving this GRE Permutation Question: Compute the number of possibilities

  1. If the student entered through A, she could leave through B or C or D or E.
    There are 4 possibilities of entering through A and leaving through a different door viz., AB, AC, AD, AE.
  2. If the student entered through B, she could leave through A or C or D or E.
    There are 4 possibilities of entering through B and leaving through a different door viz., BA, BC, BD, BE.
  3. If the student entered through C, she could leave through A or B or D or E.
    There are 4 possibilities of entering through C and leaving through a different door viz., CA, CB, CD, CE.
  4. If the student entered through D, she could leave through A or B or C or E.
    There are 4 possibilities of entering through D and leaving through a different door viz., DA, DB, DC, DE.
  5. If the student entered through E, she could leave through A or B or C or D.
    There are 4 possibilities of entering through E and leaving through a different door viz., EA, EB, EC, ED.

There are a total of 4 + 4 + 4 + 4 + 4 = 20 ways.
We could also arrive at the same answer by saying - there are 5 ways to enter AND 4 ways to leave.
So, a total of 5 × 4 = 20 ways.

Alternative ways to express the answer

The answer to this question may also be expressed in alternative ways. Here are couple of those and the rational behind those expressions.
Ask yourself the following two questions

Q1. Sampling with replacement or without replacement?

Is the problem in hand one of sampling with replacement or sampling without replacement
If the outcome of the first sampling is not put back into the sample space, we are dealing with sampling without replacement.

An effective way of determining will be to look at the choices available for each of the sampling. There were 5 choices to select a door to enter. But we were left with only 4 to exit. So, the door selected to enter is no longer available in the sample space when we had to select a door to exit.

If we had 'n' choices for the first sampling and it reduces to [n -1] and further down to [n -2] and so on, we are dealing with sampling without replacement. So, without doubt this question is an example of "sampling without replacement".

If the question is a "sampling without replacement question", the number of ways of selecting r objects from n objects is nCr.
So your first step is to compute this value. For this question, we can select 2 doors out of 5 in 5C2 ways.

Q2. Whether or not order matters in this sampling?

One of the outcomes was selecting doors AB. Another one of the outcomes was selecting doors BA.
Are these two choices the same or are they different?

  • If AB is the same as BA, then order does not matter.
  • If AB is not the same as BA, then order matters.

In this question, entering by A and leaving by B is entirely different from entering by B and leaving by A. Therefore, AB is not the same as BA. So, order matters.

If order matters, multiply the answer you got in the previous step with the number of ways things can be reordered
r distinct objects can be reordered in r! ways.
Therefore, total outcomes = 5C2 × 2!

The same may also be expressed as 5P2
nPr = nCr × r!

Choice C is the correct answer

Who would’ve thought that an old TV game show could inspire a statistical problem that has tripped up mathematicians and statisticians with Ph.Ds? The Monty Hall problem has confused people for decades. In the game show, Let’s Make a Deal, Monty Hall asks you to guess which closed door a prize is behind. The answer is so puzzling that people often refuse to accept it! The problem occurs because our statistical assumptions are incorrect.

The Monty Hall problem’s baffling solution reminds me of optical illusions where you find it hard to disbelieve your eyes. For the Monty Hall problem, it’s hard to disbelieve your common sense solution even though it is incorrect!

The comparison to optical illusions is apt. Even though I accept that square A and square B are the same color, it just doesn’t seem to be true. Optical illusions remain deceiving even after you understand the truth because your brain’s assessment of the visual data is operating under a false assumption about the image.

I consider the Monty Hall problem to be a statistical illusion. This statistical illusion occurs because your brain’s process for evaluating probabilities in the Monty Hall problem is based on a false assumption. Similar to optical illusions, the illusion can seem more real than the actual answer.

To see through this statistical illusion, we need to carefully break down the Monty Hall problem and identify where we’re making incorrect assumptions. This process emphasizes how crucial it is to check that you’re satisfying the assumptions of a statistical analysis before trusting the results.

Monty Hall asks you to choose one of three doors. One of the doors hides a prize and the other two doors have no prize. You state out loud which door you pick, but you don’t open it right away.

Monty opens one of the other two doors, and there is no prize behind it.

At this moment, there are two closed doors, one of which you picked.

The prize is behind one of the closed doors, but you don’t know which one.

Monty asks you, “Do you want to switch doors?”

The majority of people assume that both doors are equally like to have the prize. It appears like the door you chose has a 50/50 chance. Because there is no perceived reason to change, most stick with their initial choice.

Time to shatter this illusion with the truth! If you switch doors, you double your probability of winning!

What!?

How to Solve the Monty Hall problem

When Marilyn vos Savant was asked this question in her Parade magazine column, she gave the correct answer that you should switch doors to have a 66% chance of winning. Her answer was so unbelievable that she received thousands of incredulous letters from readers, many with Ph.D.s! Paul Erdős, a noted mathematician, was swayed only after observing a computer simulation.

It’ll probably be hard for me to illustrate the truth of this solution, right? That turns out to be the easy part. I can show you in the short table below. You just need to be able to count to 6!

It turns out that there are only nine different combinations of choices and outcomes. Therefore, I can just show them all to you and we calculate the percentage for each outcome.

You Pick Prize Door Don’t Switch Switch
1 1 Win Lose
1 2 Lose Win
1 3 Lose Win
2 1 Lose Win
2 2 Win Lose
2 3 Lose Win
3 1 Lose Win
3 2 Lose Win
3 3 Win Lose
3 Wins [33%] 6 Wins [66%]

Here’s how you read the table of outcomes for the Monty Hall problem. Each row shows a different combination of initial door choice, where the prize is located, and the outcomes for when you “Don’t Switch” and “Switch.” Keep in mind that if your initial choice is incorrect, Monty will open the remaining door that does not have the prize.

The first row shows the scenario where you pick door 1 initially and the prize is behind door 1. Because neither closed door has the prize, Monty is free to open either and the result is the same. For this scenario, if you switch you lose; or, if you stick with your original choice, you win.

For the second row, you pick door 1 and the prize is behind door 2. Monty can only open door 3 because otherwise he reveals the prize behind door 2. If you switch from door 1 to door 2, you win. If you stay with door 1, you lose.

The table shows all of the potential situations. We just need to count up the number of wins for each door strategy. The final row shows the total wins and it confirms that you win twice as often when you take up Monty on his offer to switch doors.

Why the Monty Hall Solution Hurts Your Brain

I hope this empirical illustration convinces you that the probability of winning doubles when you switch doors. The tough part is to understand why this happens!

To understand the solution, you first need to understand why your brain is screaming the incorrect solution that it is 50/50. Our brains are using incorrect statistical assumptions for this problem and that’s why we can’t trust our answer.

Typically, we think of probabilities for independent, random events. Flipping a coin is a good example. The probability of a heads is 0.5 and we obtain that simply by dividing the specific outcome by the total number of outcomes. That’s why it feels so right that the final two doors each have a probability of 0.5.

However, for this method to produce the correct answer, the process you are studying must be random and have probabilities that do not change. Unfortunately, the Monty Hall problem does not satisfy either requirement.

Related post: How Probability Theory Can Help You Find More Four-Leaf Clovers

How the Monty Hall Problem Violates the Randomness Assumption

The only random portion of the process is your first choice. When you pick one of the three doors, you truly have a 0.33 probability of picking the correct door. The “Don’t Switch” column in the table verifies this by showing you’ll win 33% of the time if you stick with your initial random choice.

The process stops being random when Monty Hall uses his insider knowledge about the prize’s location. It’s easiest to understand if you think about it from Monty’s point-of-view. When it’s time for him to open a door, there are two doors he can open. If he chose the door using a random process, he’d do something like flip a coin.

However, Monty is constrained because he doesn’t want to reveal the prize. Monty very carefully opens only a door that does not contain the prize. The end result is that the door he doesn’t show you, and lets you switch to, has a higher probability of containing the prize. That’s how the process is neither random nor has constant probabilities.

Here’s how it works.

The probability that your initial door choice is wrong is 0.66. The following sequence is totally deterministic when you choose the wrong door. Therefore, it happens 66% of the time:

  1. You pick the incorrect door by random chance. The prize is behind one of the other two doors.
  2. Monty knows the prize location. He opens the only door available to him that does not have the prize.
  3. By the process of elimination, the prize must be behind the door that he does not open.

Because this process occurs 66% of the time and because it always ends with the prize behind the door that Monty allows you to switch to, the “Switch To” door must have the prize 66% of the time. That matches the table!

Related post: Luck and Statistics: Do You Feel Lucky, Punk?

If Your Assumptions Aren’t Correct, You Can’t Trust the Results

The solution to Monty Hall problem seems weird because our mental assumptions for solving the problem do not match the actual process. Our mental assumptions were based on independent, random events. However, Monty knows the prize location and uses this knowledge to affect the outcomes in a non-random fashion. Once you understand how Monty uses his knowledge to pick a door, the results make sense.

Ensuring that your assumptions are correct is a common task in statistical analyses. If you don’t meet the required assumptions, you can’t trust the results. This includes things like checking the residual plots in regression analysis, assessing the distribution of your data, and even how you collected your data.

For more on this problem, read my follow up post: Revisiting the Monty Hall Problem with Hypothesis Testing.

As for the Monty Hall problem, don’t fret, even expert mathematicians fell victim to this statistical illusion! Learn more about the Fundamentals of Probabilities.

To learn about another probability puzzler, read my post about answering the birthday problem in statistics!

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