can someone please explain..
is it 60?

arjunshn wrote:How many 5 digit palindromes can be made using 1,2,3?

as you didn't specify what is palindrome in your question..so i will first of all like to give a bit definition of it..
A palindrome is a word, phrase, number or other sequence of units that can be read the same way in either direction..!!

now coming to your number..!!!
case 1)
no. of palindromes that consist only one word e.g. 11111 would be 3, (other 2 would be 22222,33333);
case 2)
- - 1 - -; now first two and last two spaces must be same..;
21112;
12121;
13131;
31113;
23132;
32123;
- - 2 - -;
21212;
12221;
13231;
31213;
23232;
32223;
- - 3 - -;
21312;
12321;
13331;
31313;
23332;
32323;

hence total no. of palindrome would be 3+6+6+6=21;

arjunshn wrote:How many 5 digit palindromes can be made using 1,2,3?

Not a real GMAT question (on the GMAT, "palindrome" would be defined, there would be more specificity regarding repeating numbers and, of course, there would be answer choices), but an interesting counting question nonetheless.

Ignoring the actual numbers, let's use X, Y and Z to represent the 3 available digits and see how many patterns we can create.

XXXXX
(1 pattern * 3, since we can replace X with Y or Z, = 3 patterns)

XXYXX
XXZXX
(2 patterns * 3, since we can replace X with Y or Z and Y or Z with X, = 6 patterns)

XYYYX
XZZZX
XYZYX
XYXYX
XZYZX
XZXZX
(6 patterns *3, since we can replace X with Y or Z and Y or Z with X, = 18 patterns)

3 + 6 + 18 = 27 possible palindromes.

(manpsingh87 forgot about the second set of patterns above.)

Stuart Kovinsky wrote:

arjunshn wrote:How many 5 digit palindromes can be made using 1,2,3?

Not a real GMAT question (on the GMAT, "palindrome" would be defined, there would be more specificity regarding repeating numbers and, of course, there would be answer choices), but an interesting counting question nonetheless.

Ignoring the actual numbers, let's use X, Y and Z to represent the 3 available digits and see how many patterns we can create.

XXXXX
(1 pattern * 3, since we can replace X with Y or Z, = 3 patterns)

XXYXX
XXZXX
(2 patterns * 3, since we can replace X with Y or Z and Y or Z with X, = 6 patterns)

XYYYX
XZZZX
XYZYX
XYXYX
XZYZX
XZXZX
(6 patterns *3, since we can replace X with Y or Z and Y or Z with X, = 18 patterns)

3 + 6 + 18 = 27 possible palindromes.

(manpsingh87 forgot about the second set of patterns above.)

thanks a lot sir....!!!

arjunshn wrote:How many 5 digit palindromes can be made using 1,2,3?

Using 1,2 and 3 only, we need to count how many ways we can build a 5-digit integer that is the same backwards and forwards, such as 12221, 32123, 22122, and so on.

Ten-thousands digit can be 1,2, or 3 = 3 choices.
Units digit must be the same as the ten-thousands digit = 1 choice.
Thousands digit can be 1,2,or 3 = 3 choices.
Tens digit must be the same as the thousands digit = 1 choice.
Hundreds digit can be 1,2, or 3 = 3 choices.

To combine the number of choices for each digit, we multiply:
3*1*3*1*3 = 27.

The key to solving such a question is to identify that we have to chose only three digits,

EITHER We chose the digits at the units, tens and the Hundredth place or we chose digits at the ten thousandth, thousandth and hundredth place -
Each of the digits can be chosen in 3 ways - i.e. the digits -1 or 2 or 3



thus the answer should be 3*3*3 = 27

Using 1,2 and 3 only, we need to count how many ways we can build a 5-digit integer that is the same backwards and forwards, such as 12221, 32123, 22122, and so on.

Ten-thousands digit can be 1,2, or 3 = 3 choices.
Units digit must be the same as the ten-thousands digit = 1 choice.
Thousands digit can be 1,2,or 3 = 3 choices.
Tens digit must be the same as the thousands digit = 1 choice.
Hundreds digit can be 1,2, or 3 = 3 choices.

To combine the number of choices for each digit, we multiply:
3*1*3*1*3 = 27.

Thanks ! this is an awesome solution !