Amy is organizing her bookshelves and finds that she has

This looks like a pattern question. i tried to just understand what they were asking and picked out some letters to test:

if 1 letter, only allowed 1 code (i.e. A)

if 2 letters, allowed 3 codes (i.e. A, B, AB)

if 3 letters, allowed 6 codes (i.e. A, B, C, AB, AC, BC)

if 4 letters, allowed 10 codes (i.e. A, B, C, D, AB, AC, AD, BC, BD, CD)

kudos if helpful

Hi Cinematiccuisine!

Happy to help

Since the numbers we're dealing with here are small, it is probably easiest to just solve this manually, at first, at least to understand what the question is asking about. Amy is labeling each type of book with either one or two letters. If she is using just one letter, this means there is only one code she can use:

A

If she is using two letters, then there are three:

A
B
AB

We are told that the order of the letters doesn't matter, so we don't have to worry about counting AB and BA separately. Now, with three letters, we have all of the above combinations, plus three more:

C
AC
BC

That's six total. If we add one more, then we have those six plus four more:

D
AD
BD
CD

And that give us 10 possible codes, which is what the question is asking for So the answer is 4.

Now, if the number was much bigger, then we might need to create a formula to solve this. The easiest way to do that is to just notice the pattern. Every time we add a new letter, we get one more code (just that letter), plus the number of letters that we already have, which create the combinations of mixed letters (like AD, BD, CD). So adding C gave us 3 more codes, adding D gives us 4 more codes, adding E gives us 5 more codes, and so on. We can use that pattern to determine the number of possible codes at pretty much any step

We can write this out in terms of combinations, too. If N is the number of letters that we're using, then the number of codes can be written as:

N + NC2

So with four letters, we get 4 + 4C2 = 10 different codes.

I hope that helps!
-Carolyn

Please put the first few words as the topic name.

What it means?
10 books have to be given a code as a single letter such as A, B etc or as a double letter such as AB, BA, AC etc. AB and BA will be different as both are different as code

Solution
Let the number of letters required is n,
So single digits will be n
Two digits will be nC2........
.we are not multiplying with 2! because order does not matter.
So \(n+nC2\geq{10}\)........\(n+\frac{n!}{(n-2)!2!}\geq{10}........n+n(n-1)/2\geq{10}...............(2n+n^2-n}\geq{20}............n(n+1)\geq{20}.........n(n+1)\geq{4*5}..\)
So minimum value = 4

B

Solution:

Since there are only 10 books, the number of colors needed is even fewer. Let’s check the answer choices starting with 3.

If there are 3 colors, the number of single-color codes is 3 and the number of codes that are pairs of two different colors is 3C2 = (3 x 2)/2 = 3. Therefore, if there are 5 colors, the total number of codes can be formed is 3 + 3 = 6. However, this is not sufficient since there are 10 books.

If there are 4 colors, the number of single-color codes is 4 and the number of codes that are pairs of two different colors is 4C2 = (4 x 3)/2 = 6. Therefore, if there are 4 colors, the total number of codes can be formed is 4 + 6 = 10. We see that this is sufficient since there are exactly 10 books.

Answer: B

I wrote all possible answer choices down:
A
AB (since Order does not matter we will not take BA into account)
B
CA
CB
C
DA
DB
DC
D

Fairly straightforward question. No reason for it to not be "GMAT style"

a) Three letters give : 3 + 3C2 = 3 + 3 = 6

b) Four letters give : 4 + 4C2 = 4 + 12 = 16

So, answer is B

Hello - Just a quick thing, the question says "the order of the letter does not matter", so we should be using combinations and permutations.

-----------------------
Kudos, if my post helped

three letters could be enough for uniquely naming 9 books i.e (3! *3) and remaining 1 could be named by taking one more letter under consideration.
Least total letters would be 10

Amy is organizing her bookshelves and finds that she has 10 different types of books. She then codes each book with either a single letter or a pair of two different letters. If each type of book is uniquely represented by either a single letter or pair of letters, what is the smallest number of letters Amy will need to create the codes for all 10 types of books? (Assume the order of letters in a pair does not matter.)

A) 3
B) 4
C) 5
F) 10
E) 20

Questions:
1) How do i rephrase this question? The Kaplan answer is not so easy to understand ( at least for me)
2) without getting into too many calculations, how can i solve this?

Thank you for your help.

Merging topics.

Please follow the rules when posting a question: https://gmatclub.com/forum/rules-for-po ... 33935.html

The question asks what is the smallest number of letters required to create at least 10 unique codes. Because the question told us that the books are numbered by EITHER a single letter or 2 different letters, and we need to find the least number of letters to create 10 codes, we can safely look into the situation in which all books are numbered by 2 different letters.

Let's call "n" is the number of letters we need to find, then we have

n x n(-1) >= 10
4 x 3 =12 -> MORE than 10 -> keep this
3 x 2 = 6 -> LESS than 10 -> discard
5 x 4 = 20 -> MORE than 10 -> but n=5 < 4 -> we pick the n=4

For these type of questions, let us assume that n distinct letters/digits are required.

then for second code of two digits numbers can be selected as nC2

thus we get total number of options :
10 = n+ nC2
=> 2n + n(n-1) = 20
=> n^2-n-20=0
=> (n-5)(n+4)=0
or n=5 as n cannot be negative.

total letters required ;
a,b,c,d; 4
pair can be made
ab,ac,ad,bc,bd,cd,a,b,c,d
IMO B

Answer is B.

A
B
AB
C
AC
BC
BA
CA
CB
D

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