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# Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address? GMAT Explanation with strategies and more examples!

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8

Good old Tanya’s letters! This is from the GMAT Prep Tests 1 and 2 so if you haven’t tackled those then hold off on working through this.

## Tanya and her letters

This is a tricky GMAT combinatorics probability question that has a somewhat simple solution. And that’s something that you need to keep in mind on the GMAT. Not everything is about elegance. Or math power. Or some trick or strategy. A lot of your success on the GMAT will rely on making step by step, organized, practical choices. Not all combinatorics questions need to be tackled with factorials or the slot method.

Some questions are best solved by writing out scenarios and/or spotting a pattern. Tanya and her letters is one such question. It can be solved purely with combinatorics formulas but most people will probably be more successful at least doing part of this one with a more practical approach.

So what’s a sign for: don’t cram it into a formula and stay practical! The biggest sign, there aren’t that many possibilities. So writing it out wouldn’t be a big time suck (there are only 4 envelopes and 4 letters). Secondarily, if the setup feels non-standard or has a bunch of moving parts that don’t fit well into a formula that might indicate that a more practical/non-math setup would make more sense.

## Let’s start by defining the question

what is the probability that only 1 letter will be put into the envelope with its correct address?

OK – no problem. Probability doesn’t mean difficult. We need, Our Scenario/Total. So, only 1 letter will be put into the envelope with its correct address/Total possibilities.

## Now go ahead and start setting things up/organizing information

Letters: A, B, C, D

Addresses/Envelopes: a, b, c, d (I used lowercase versions of the letters to make it easy to match up correct letters/addresses)

The total is usually the easiest thing to figure out because you likely won’t have any constraints that could make the setup trickier. In this case we have 4 options for letters and 4 options for addresses. So you have 4 spaces for letters (treat this as an ordering question). How many different ways can you arrange 4 things? 4! = 24.

Now we need to calculate the numerator which is the specific scenario: only 1 letter will be put into the envelope with its correct address

For this you might just write it out. It’s quick. You should see the pattern after the first two so you don’t have to write out the rest.

Aa Bd Cb Dc

Aa Bc Cd Db

Bb Ac Cd Da

Cc Ab Bd Db

Dd Ac Ba Cb

Dd Ab Bc Ca

## You can also do this with a formula which is neater but not necessarily faster or easier/less prone to error.

There are 4 correct matchups:

Aa

Bb

Cc

Dd

And then you need to figure out the remaining incorrect ones. So, let’s just take the first correct one Aa. For the next slot, b you have 2 options (c, d) but once you’ve chosen c or d you don’t have choices for the last two slots, you’re locked in, you only 1 option each because any other arrangement will create a “correct” address/letter combination.

So you have 4 correct options (Aa, Bb, Cc, Dd) and then 2 options for the incorrect ones. 4*2 = 8.

That’s not complicated but I see a lot of GMAT tutoring students get tangled up there. And considering you don’t have to write out much to see the pattern, a practical approach is usually more successful on this question.

## Finally, let’s put it all together

8/24 = 1/3.

So 1/3 of the time you are left with only one of Tanya’s letters in the envelope with the correct address.

## More GMAT Combinatorics and Probability Questions!

Here’s a great one from the GMAT official guide that also touches on the idea of being practical with combinatorics: A researcher plans to identify each participant in a certain medical experiment

And another tough oddball combinatorics question from the GMAT Prep Tests: A certain office supply store stocks 2 sizes of self-stick notepads, each in 4 colors

And here are two tough official GMAT probability questions:

A basket contains 5 apples, of which 1 is spoiled and the rest are good.

If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of , where b is an integer?

# Among a group of 2,500 people, 35 percent invest in municipal bonds, 18 percent invest in oil stocks, and 7 percent invest in both municipal bonds and oil stocks. If 1 person is to be randomly selected from the 2,500 people, what is the probability that the person selected will be one who invests in municipal bonds but NOT in oil stocks? GMAT Explanation

Among a group of 2,500 people, 35 percent invest in municipal bonds, 18 percent invest in oil stocks, and 7 percent invest in both municipal bonds and oil stocks. If 1 person is to be randomly selected from the 2,500 people, what is the probability that the person selected will be one who invests in municipal bonds but NOT in oil stocks?

(A) 9/50
(B) 7/25
(C) 7/20
(D) 21/50
(E) 27/50

Combinatorics questions, especially ones tangled up with probability, scare students. Can there be a very challenging probability question? Yes! Are most of them tough? Not really. Do you need a bunch of complex math to solve GMAT probability questions? Nope. Let’s see what we need to get this one solved.

## Define the question

What is the probability that the person selected will be one who invests in municipal bonds but NOT in oil stocks?

OK, not so bad. Munis but no oil stocks. So what is probability exactly? Specific Scenario/Total Scenarios. So Muni/Not Oil/Total. This is starting to feel like overlapping sets with a side of probability…

## Let’s get the information organized!

Let’s call the 2500 people 100 people. It won’t make a difference in the answer because everything is based on percents/proportions and will make it easier to do the calculations. So:

Municipal Bonds: 35

Oil Stocks: 18

Muni and Oil: 7

Looks like we have a two group overlapping set. We’re going to need a diagram.

So just fill in the given information and then make easy inferences. You’ll be able to get the whole chart filled in. You end up with 28 people that have municipal bonds but not oil stocks. But, 28 isn’t the answer because we’re asked about probability. So take 28 and put it over the total. What’s the total? The 2500. BUT we changed that to 100. 28/100 = 7/25.

## More GMAT two group overlapping sets and probability practice!

Here’s a challenging sets question from the GMAT Prep Tests 1 and 2. Stay organized. Make the easy inferences and you should be fine.

In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:

Here’s one from GMAT Question of the Day that’s very similar in that the answer is based on proportions. It doesn’t have a probability component but again: same idea and great practice.

GMAT Question of the Day 2 group overlapping sets proportions and percents

This one is a bit different but will still give you 2 group practice and will give you exposure to overlapping sets on data sufficiency which often comes down to counting equations.

GMAT Question of the Day 2 group overlapping sets Data Sufficiency

And here’s a very tough DS overlapping sets questions. Again, deals with counting equations/system of questions.

GMAT Question of the Day Challenging 2 Group Overlapping Sets Data Sufficiency Counting Equations

# A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code? GMAT Explanation, Video Solution, and Additional Practice Questions!

A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4

B. 5

C. 6

D. 7

E. 8

We need to help this researcher identify 12 participants using single letters or pairs of letters in alphabetical order. Key info: we want the LEAST number of letters to make these 12 codes. The word code should trigger: ordering! Yes, codes usually involve an ordering element. However, not always and this is one of the exceptions. Whenever a code has to be in alphabetical order or ascending/descending order it is highly likely that it no longer has an ordering element and should be treated as a group.

So what does that mean for our researcher and his medial experiment? Well, let’s start testing the answers to find the least value that can create 12 codes. With that in mind start with the smallest number. Why? Because if the smallest number works then it’s the correct answer. If say, 6 works, then you still need to test 5 and potentially 4. If looking for a max you might start with the biggest answer choice.

Ok, so how do we test 4? Well, you have 4 single digit codes say A, B, C, D.

But how about the pairs? Again, think of this as a group. So it’s a group of 2 with 4 choices. Or 4 choose 2.

4*3/1*2 = 6

So 4 (single) + 6 (pairs) = 10.

Not enough to identify our twelve participants!

Let’s try 5.

A, B, C, D, E

5 single codes.

5*4/1*2 = 10

10 pairs

10 + 5 = 15. That works!

So the least number of letters needed to identify the 12 participants: 5. B.

That’s one way to do it. There’s also a very practical way to approach this combinatorics puzzle. And, all over the GMAT, especially on combinatorics questions with a very limited numbers of possibilities (we’re only looking for 12 codes here) remember to stay down to earth. Sometimes you can just make a list. And sometimes that will be the fastest approach. Again: stay practical on the GMAT!

So let’s try 4 letters.

Single Letter Codes: 4

A, B, C, D

Pairs: 6

AB. BC. DE

AC. BD

Total Codes: 10

Let’s try 5 letters.

Single Letter Codes: 5

A, B, C, D, E

Pairs: 10

AB  BC. CD. DE

AC. BD. CE

AE

Total Codes: 15

If you’re taking a practical approach take your time planning. Think about how you want to make your list. Don’t rush it. What I see a lot in GMAT tutoring is students having a decent understanding of how to approach things from a practical perspective BUT because they’re not using a formula or test prep technique don’t feel confident and then rush the setup/execution. Then they get the feeling that the practical approach doesn’t work and that they need a formula for everything. So, again, if you’re going for a non-formulaic setup, which sometimes is exactly what you should do, give it the space to succeed.

## Additional GMAT Combinatorics Practice Question Examples

Here is a grouping question from the GMAT Prep Tests. This is using the same method of grouping as the question above (although it has different conditions/constraints).

Here is a somewhat standard GMAT Combinatorics Code example from GMAT Question of the Day

Finally another wordy combinatorics example from question of the day

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?

(A) 3/7
(B) 5/12
(C) 27/70
(D) 2/7
(E) 9/35

## Additional GMAT Combinatorics/Grouping Practice Questions

Here’s another grouping question from the GMAT Official Guide: A researcher plans to identify each participant in an experiment

# A certain office supply store stocks 2 sizes of self-stick notepads, each in 4 colors: Blue, Green, Yellow or Pink. The store packs the notepads in packages that contain either 3 notepads of the same size and the same color or 3 notepads of the same size and of 3 different colors. If the order in which the colors are packed is not considered, how many different packages of the types described above are possible?

(A) 6
(B) 8
(C) 16
(D) 24
(E) 32

Here’s a challenging combinatorics question from the GMAT Prep Tests. We almost always see the self-stick notepad question come up in the lifecycle of a GMAT prep here at Atlantic. There’s not much work to do as the office supply store only stocks two sizes and four colors (limited options should signal the possibility of just writing out the scenarios). Keep in mind that most of this question (and many GMAT questions) relies on simply understanding the text and then getting the information organized.

Yes, it is helpful to have some basic combinatorics tools. But that aspect of this question is really lightweight as you can see in the diagram below.

## Office Supply Store Self-Stick Notepads Solution!

The first step is to figure out the number of self-stick notepads that have the same size and same color. Set up one slot for size and the other for color. Then do the same things for same size but different color. This part is a little tricker as you need to consider the color aspect more specifically.

For the first pad you can choose all 4 colors. But for the second, because you can’t use the same color that you chose for the first notepad, you can only choose 3 colors. For the third pad you can choose 2 colors. You need a denominator for the colors because you are making a group (order doesn’t matter). So you need to divide out the duplicates (green, blue, yellow is the same as blue, green, yellow so you shouldn’t count both of those).

Because you don’t have that many options this could be a question on which you might avoid formulas altogether and just write out the scenarios or do some combination of practical and formula.

Here is the solution with the slot method described above:

## Here’s a practical approach to counting these self-stick notepads

Different Size – Same Color

Size 1 notepads in 4 colors (all blue, all green, all yellow, or all pink)  = 4 options

Size 2 notepads in 4 colors = same as size 1 = 4 options

Total = 8

Different Size – Different Color

Size 1 ABC, ABD, ACD, CBD

Size 2 SAME COLOR POSSIBILITIES AS SIZE 1

Total = 8

8 + 8 = 16.

## Here’s another multi-step challenging GMAT Combinatorics question with an in depth explanation:

Tanya prepared 4 different letters to be sent to 4 different addresses.

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