A palindrome is a number that reads the same forward and bac

XYZYX

X can be 1, 2, or 3, thus 3 options.
Y can be 1, 2, or 3, thus 3 options.
Z can be 1, 2, or 3, thus 3 options.

Total 3^3=27.

Answer: E.

Similar questions to practice:
a-palindrome-is-a-number-that-reads-the-same-forward-and-129898.html
a-palindrome-number-reads-the-same-backward-and-forward-159265.html

Hope this helps.
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There are 3 options for the 1st digit, so this implies that there is 1 option for the 5th digit, because it must be the same as the 1st digit.
There are 3 options for the 2nd digit, so this implies that there is 1 option for the 4th digit, because it must be the same as the 2nd digit.
There are 3 options for the 3rd digit (because it is the middle).

3*3*3*1*1 = 27 options

Take the task of creating a 5-digit palindrome and break it into stages.

Stage 1: Select a digit for the first position.
We can choose 1, 2 or 3, so we can complete stage 1 in 3 ways

Stage 2: Select a digit for the second position.
We can choose 1, 2 or 3, so we can complete stage 2 in 3 ways

Stage 3: Select a digit for the third position.
We can choose 1, 2 or 3, so we can complete stage 3 in 3 ways

Stage 4: Select a digit for the fourth position.
Important: In order to create a palindrome, the fourth digit must be the same as the second digit.
For example, if the first three digits are 213, then fourth digit must be 1, and the fifth digit must be 2 to get the 5-digit palindrome 21312
Since the fourth digit must be the same as the second digit, we can complete stage 4 in 1 way

Stage 5: Select a digit for the fifth position.
Since the fifth digit must be the same as the first digit, we can complete stage 5 in 1 way

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus create a 5-digit palindrome) in (3)(3)(3)(1)(1) ways (= 27 ways)

Answer: E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Started with 1:

12321
13231
12221
13331
11111
11311
11211
12121
13131

9 possibilities for 1, 9 for 2, and 9 for 3 =

Similar questions to practice:
https://gmatclub.com/forum/a-palindrome- ... 29898.html
https://gmatclub.com/forum/a-palindrome- ... 59265.html

Hope this helps.
_________________

Hi Bunuel,

Can you please explain the logic behind the same? There are many kinds of palindromes that can be formed using two or three numbers (e.g. 12221, 12121, 12321, 32123) I am unable to understand how your approach caters to all of them?

Please help.

Thank you

Regards,
Neha

In a five-digit palindrome XYZYX we are concerned about X, Y and Z, so about the first three digits.

You get 12221 if you choose 1, for X, 2 for Y and 2 for Z, the same for other options.
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Voila! Makes complete sense, thank you so much for responding!




In a five-digit palindrome XYZYX we are concerned about X, Y and Z, so about the first three digits.

You get 12221 if you choose 1, for X, 2 for Y and 2 for Z, the same for other options.[/quote]

I did the counting method. Was able to do it under 2 minutes.
Since it is a palindrome, I only counted the first 3 digits since the last 2 would mirror the initial 2 digits.

- All possible permutations of XYZ (All digits different) --> 6
- All possible permutations of XXZ (1st & 2nd Digit are the same) --> 6
- All possible permutations of XZZ (1st & 3rd Digit are the same) --> 6
- All possible permutations of XYX (1st & 3rd Digit are the same) --> 6
- All possible permutations of XXX (All digits same) --> 3

n=6+6+6+6+3=27 --E

I can choose out of (1,2 and 3) for the first value.
I can choose out of (1,2 and 3) for the second value and
I can also choose out of (1,2 and 3) for the third value.

However: To get a "palindrome" number the two proceeding terms have to read the same as the first two. As I have already chosen values for the first and the second term I logically have only one choice for the next two, which is the one that I already chose.

Therefore it is: 3 * 3 * 3 * 1 * 1 = 27 possible combinations

Since this is a 5 digit palindrome, the middle number would act like a mirror.

Case 1: Middle number = 1
_ _ 1 _ _
Number of options for 1st place = 3
Number of options for 2nd place = 3
Number of options for 4th and 5th place = 1
Total ways = 3*3 = 9

Case 2: Middle number = 2
_ _ 2 _ _
Similar to the above reasoning,
Total ways = 3*3 = 9

Case 3: Middle number = 3
_ _ 3 _ _
Similar to the above reasoning,
Total ways = 3*3 = 9

Total number of ways = 9 + 9 + 9 = 27

Correct Option: E

Let us take one case

1---1

The inside 3 digits can be formed with one digit occurring thrice i.e, 111, 222, 333 in 3 ways
And with one digit occurring once i.e, 121, 131, 212, 232, 313, 323 in 6 ways

So there are 9 ways with 1 at the ends

Totally 27 ways considering 1, 2 and 3
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_ _ fist two numbers

1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3

9 options


_ middle number

we can change this number, thus 3 options




_ _ _ last three numbers

they correspond to the first 3 numbers

1 option



9*3*1 = 27




therefore 5*3*1=27
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We have 3 options for the first digit, 3 options for the second, 3 options for the third, 1 option for the fourth (since it has to be the same as the second), and 1 option for the fifth (since it has to be the same as the first). Thus, there are 3 x 3 x 3 = 27 possible palindromes.

Answer: E
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Take a very easy approach here...

5 digit palindrome of ABCDE being 5 digits
and by definition since should be the same backwards A=E, B=D
therefore, we get ABCBA as 5 digits

options being 1,2,3

Therefore A can be =1,2,3 ( 3 options)
B can also be = 1,2,3 (3 Options)
C can be 1,2,3 (3 options)

=> 3 x 3 x3 = Total number of outcomes possible. = 27 Answer E

Had we not been allowed to use a digit more than twice, then our outcome could have been 3 x 2 x 1 = 6 (Maybe) please comment your views

Hello

I get the idea while counting the possibilities....
However if we want to coin the concept or formula.. could we say that it is like distributing different objects to different bins? ( the term coined to a certain type of combination problems)
so we say
3 digits put in 3 bins... we get same result

Hello! Please let me know if this is correct:

CASE 1: All same numbers (use just one digit)

Ex. 11111

Possibilities = 3

Listing it: 11111, 22222, 33333

CASE 2: Alternating 2 numbers (use two digits)

Ex. 12121

Possibilities = 6

3 x 2 x 1 x 1 x 1

Listing it: 12121, 21212, 13131, 31313, 23232, 32323

CASE 3: Alternating three numbers (use three digits)

Ex. 32123

Possibilities = 18

3 x 2 x 1 x 1 x 1 TIMES # of ways they can switch which is 3

Listing it: 12321, 32123, 21312, and so on

Add all possibilities --> 3 + 6 + 18 = 27

1st 2nd 3rd 4th 5th
___ ___ ___ ___ ____

The 1st place can be filled by (1,2,3)= \(3 \)ways

also, The 2nd and 3rd place can be filled by (1,2,3)=\(3\) ways(for both 2nd and 3rd place)

the 4th place digit has to be same with the digit in 2nd place and 5th place digit has to be same with the digit in 1st place to make a palindrome =1 ways (for both 4th and 5th place)

total =\(3*3*3*1*1=27 \)

Answer E

Is this a correct method?

The ways in which we can rearrange either 1, 2 or 3:
(5!)/(3!2!)

Multiply by 3:

(5!)/(3!2!) * 3

And then subtract 3 because we need to then subtract the instances where we get either 111, 222 or 333.

(5!)/(3!2!) * 3 - 3

= 27