# Beat The GMAT Math Challenge Question – October 4, 2010

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A palindrome is a word that is read the same backwards as forwards. For example, the words “BADAB,” “IAGAI,” and “HHHHH” are all palindromes.

How many 5-letter palindromes can be created using the letters A, B, C, D, E, F, G, H, I and J?

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## 37 comments

Abhishek Pandey on October 4th, 2010 at 6:26 am

Hi all,

I hope the answer is 1000

The explanation is as follows:

We need to create five letter palindrome

Number of digits we have is - A, B, C, D, E, F, G, H, I, J = 10 digits

To get 5 letter palindromes, we need to use the pattern xyzyx

Pattern xy can be any combination of 2 digits using the above mentioned digits ex; AA, AB, BA, CD, DC , ...

The number of such possible 'xy' outcomes is 10 * 10 = 100

The value z can be any of the 10 digits ex- A, B, C, D, E, ...

The number of such possible 'z' outcomes is 10

So total number of 5 letter palindromes are -- 100 * 10 = 1000

One easy formula --

Suppose we are using all the alphabets

=> If we need 'n' letter palindrome, and n is even -- then formula - 26^(n/2)

Example: n = 6 , number of 6 letter palindromes are - 26^3

=> If we need 'n' letter palindrome, and n is odd -- then formula - 26^((n+1)/2)

Example: n = 9 , number of 9 letter palindromes are - 26^5

Kapil Bansal on October 4th, 2010 at 8:08 pm

My Answer is 1000

- - - - -

1 2 3 4 5

You can ignore 4 / 5 since they are covered by 1 and 2

Choices position 3 = 10

Choices position 2 = 10

Choices position 1 = 10

Total = 10X10X10 = 1000

Siddharth on October 4th, 2010 at 8:42 pm

I'm not sure about that formula - is there a website or source that you looked it up?

If there are 10 ways of selecting the second alphabet, then you're precluding the fact that the second alphabet could in effect be the same as the first alphabet. In this case, you're accounting for the same word in the first and second case. The same reasoning applies to case 2,3 and case 1,3 taken in pairs.

In effect, what I'm saying is that if you're accounting for AAAAA in case 1, then case 2 cannot have 10 possibilities (as it too includes AAAAA) - it can result in only 9 possibilities instead. Likewise, case 3 can have only 8 possibilities. In addition, you will need to account for different permutations of words formed by different ordering of aplhabets...I've posted an explanation in my answer below. HIH

Varun Mehta on October 4th, 2010 at 7:16 am

I suppose the answer is 2440

Chartsiri on October 4th, 2010 at 8:26 am

I think the answer is 1000

Here is my explanation:

We have 10 words - A, B, C, D, E, F, G, H, I and J- and have to create to a palindrome, as shown in the question.

We can classify into 5 scenarios:

1. AAAAA - 10

2. ABCBA - 10 x 9 x 8 = 720

3. ABBBA - 10 x 9 = 90

4. AABAA - 10 x 9 = 90

5. ABABA - 10 x 9 = 90

Therefore, the answer is 10 + 720 + 90 + 90 + 90 = 1000

Chartsiri on October 4th, 2010 at 8:32 am

I think the answer is 1000,

Here is my explanation:

5 scenarios can be classified from the question:

1. AAAAA - 10

2. ABCBA - 10x9x8=720

3. ABBBA - 10x9=90

4. AABAA - 10x9=90

5. ABABA - 10x9 =90

so, the answer is 10+720+90+90+90 = 1000

Rajesh on October 4th, 2010 at 8:53 am

Number of alphabets are 10. And we need to worry about the combinations we get for first 3 letters only for a 5 digit palindrome. So the answer is 10 * 10 * 10 = 1000.

euro on October 4th, 2010 at 9:03 am

5-letter word can be graphically represented as follows:

___ ___ ___ ___ ___

1 2 3 4 5

Total no. of letters from A to J = 10

Position 1 & 5 will have the same letters and they can have any of the 10 letters.

Similarly, position 2 & 4 will have the same letters and they can have any of the 10 letters.

Position 3 can be filled with any of the 10 available letters.

Total no. of palindromes possible = 10 x 10 x 10 = 1000 (Answer)

Anita on October 4th, 2010 at 9:46 am

hello,

I believe the answer is 3840

The explanation is-

there are 5 letters in palindrome

Number of digits we have is - A, B, C, D, E, F, G, H, I, J = 10 digits

consider first 3 letters- abc

there will be 6 letters for the first combination abc

hence with a as first letter and 'b' and second letter as combination - abc(6 letters for this combination),abd,abe,abf,abg,abh,abi,abj there are 48 letters

similarly with a as first letter and 'c' and second letter as combination

acd(6 letters for this combination),ace,acf,acg,ach,aci,acj there are 42 letters

Hence for letter a and combination as b,c,d,e,f,g,h,i,j there are 48 + 42*8= 336 i.e 384

Hence for all 10 digits the total letters would be 384*10= 3840

Nox on October 4th, 2010 at 9:52 am

To make a 5 letter palindrome, there should be <= 3 distinct individual letters and not more than that.

- 3 distinct letters: xyZyx = 10p3 = 720

- 2 distinct letters: xyXyx + xyYyx + xxYxx = 3 x 10p2 = 270

- 1 dstinct letter: xxXxx = 10p1 = 10

Adding all of the above cases, total no. of 5 letter palindromes = 1000.

Kumaran on October 4th, 2010 at 10:16 am

Answer should be 10 + 720 + 90 - 1000 ways.

X X X X X --> For all the five places a single character is chosen out of 10 characters -- 10C1 = 10 ways.

_ X X X _ --> The middle three places a single character is chosen out of 10 characters and for the first and last place one of the other 9 characters is chosen. --> 10C1 * 9C1 = 90 ways.

_ _ X _ _ --> The middle place could be chosen in 10 ways. The second and fourth in 9 ways and first and the last characters could be 8 ways. Total 10*9*8 =720 ways

Siddharth on October 4th, 2010 at 11:18 am

Hey, I guess you meant 10 + 720 + 90 = 820. Not 1000

Siddharth on October 4th, 2010 at 11:15 am

The answer is 820

A five lettered palindrome can be formed in 1 of the following 3 ways:

a) All the 5 letters are the same

b) The first two letters of the word are different and the third letter is one of the letters that comprises the first two letters

c) The first three letters are all different

My reasoning is that for a palindrome to be formed, the last 2 letters of the words are mirror images of the first two letters. Therefore, it is sufficient to deal with the mechanics of selecting the first three letters.

If all the letters of the word are same, then the number of such words are 10C1 = 10.

In the event that the first three letters of the word are different ABABA and ABBBA represent possibilities where either of the two letters could occupy the third slot. This results in 2! unique such words. Therefore, the number of such words are 10C2 * 2! = 90

In the event that the first three letters of the word are different ABCBA, ACBCA, BACAB, BCACB, CABAC and CBABC represent possibilities where either of the three letters could occupy the third slot. This results in 3! unique such words. Therefore, the number of such words are 10C3 * 3! = 720

Given this, the total number of 5 letter palindromes that can be formed is 10 + 90 + 720 = 820.

Kumaran on October 4th, 2010 at 11:32 am

@ Siddharth

Hey, I guess you meant 10 + 720 + 90 = 820. Not 1000

ARGH!!! Yes, silly addition mistake. Can you believe it .

My final answer should be be 10 + 720 + 90 = 820

Siddharth on October 4th, 2010 at 11:39 am

@Kumaran: LOL. I know its silly. Maybe the other posts above had something to do with it, just as the trap answers on the real GMAT

Srinidhi on October 4th, 2010 at 11:35 am

The Answer is 1000 ways.

Explanation: For a 5 digit palindrome,we need to consider only the first 3 digits.

First: 10 numbers possible ( A to J)

Second: 10 numbers possible ( A to J)

Third: 10 numbers possible ( A to J)

Chandra Pallaka on October 4th, 2010 at 2:09 pm

Answer is 1000.

we can repeat letters.

for 1st position - 10 letters

for 2nd position - 10 letters because no restriction on repeating letters

for 3rd position - 10 letters

for 4th position - 1 letter that was selected for 2nd position

for 5th position - 1 letter that was selected for 1st position

total is 10*10*10*1*1=1000

Sunil on October 4th, 2010 at 2:19 pm

Hi,

The answer is 1000.

Consider the digit as below:

_ _ _ _ _

Now, starting from the left, the leftmost place can be filled in 10 ways. As the word is palindrome, the rightmost place will contain the same alphabet as on the leftmost place. So number of ways of filling the leftmost and rightmost digit = 10*1 = 10

Similarly, the number of ways to fill the second place from the left and second place from the right = 10*1 = 10

Number of ways to fill the middle place = 10

So total number of ways = 10*10*10 = 1000

cm on October 4th, 2010 at 7:41 pm

aaaaa (and similar) => (1 *10) = 10

---to get aaaaa bbbbb ccccc etc = 10 ways

baaab (and similar) => (9*10) = 90

---to get baaab, caaac, daaad... abbba, cbbbc,...

babab (and similar) => (9*10) = 90

---to get babab, cacac, dadac,... ababa, acaca,...

bacab (and similar) => (8*9*10) = 720

---to get bacab, badab,...

10+ 90+90 +720 =910

Ashish on October 4th, 2010 at 9:57 pm

did you forgot to include the following case:

aabaa (and similar) = 10*9 = 90

--- to get aabaa,aacaa, aadaa,.... bbabb, bbcbb, ....

Ashish on October 4th, 2010 at 9:57 pm

did you forgot to include the following case:

aabaa (and similar) = 10*9 = 90

--- to get aabaa,aacaa, aadaa,.... bbabb, bbcbb, ....

This will take you total from 910 to 1000

Atif on October 4th, 2010 at 9:08 pm

My answers is 820.

Here is my answer with explanation;

We can use any of 10 letters if we use only one letter for all of the 5 places like AAAAA or BBBBB or CCCCC and so on till J. So, there are only 10 ways of making a palindrome if we use only one letter for all places. >>10 ways<10*9=90 ways to make a palindrome by using only 2 letters. >>90 ways<10*9*8=720 ways to ma ke a palindrome by using this combo. >>720 ways<<

If we use four different letters like ABCDA or ACDEA, then we can't make a palindrome out of 4 different letters.

So, there are 10+90+720=820 ways of making a palindrome by using given letters.

Ashish on October 4th, 2010 at 9:48 pm

IMO 1000.

12345

1 2 3 can all be in 10 ways each = 10*10*10 = 1000

4 and 5 will be restricted to either of the 100 ways depending on 1 and 2

Shubhashis on October 5th, 2010 at 12:59 am

5 letters and that to palindrome.

3rd position: does not matter what you select so it can be selected in 10 ways.

2nd position: does not matter what you select so it can be selected in 10 ways.

4th position: whatever has been selected for 2nd position only that has to be selected so it can be selected in 1 way.

1st position: does not matter what you select so it can be selected in 10 ways.

5th position: whatever has been selected for 1st position only that has to be selected so it can be selected in 1 way.

So total number of possible palindromes are 10*10*10 = 1000 .

Bibek Mishra on October 4th, 2010 at 10:48 pm

IMO answer is 1000.

1st and 5th digit have to be same and can be filled up in 10 ways

Thus 2nd and 4th digit have to be same and can be filled up in 10 ways (Because repetition of alphabets is allowed)

3rd digit can be filled in 10 ways.

So answer: 10 X 10 X 10 = 1000

jiten on October 4th, 2010 at 11:50 pm

There are two ways we can solve this problem.

1 We can choose a character (which can repeat) from the 10 letters.

First letter can be chosen in 10 ways.

Second Letter can be chosen in 10 ways.

Third = 10 ways.

Fourth & fifth position is repeat of first and second so they are fixed.

Total =10 * 10 * 10 = 1000 ways

2. We look for different combinations.

All three characters are different.

XYZXY = 10 * 9 * 8

First two are same.

XXYXX = 10 * 9

Character in Center are same

YXXXY = 10 * 9

3 Characters at odd places are also same.

XYXYX = 10 * 9

All same characters

XXXXX = 10

Total = 720 + 90 + 90 + 90 + 10 = 1000

Gurpreet singh on October 5th, 2010 at 12:53 am

The answer is 1000

Three possibilities : XXYXX, XYZYX , XXXXX

For XXXXX : 10

For XXYXX = 10*9 = 90

For XYZYX = 10*9*10 =900

as Z can take any value

Hence 10+90+900 = 1000

Khayyam on October 5th, 2010 at 3:05 am

X X X X X - 10 ways

X Y X Y X - 10*9= 90 ways

X Y Z Y X - 10*9*8 = 720 ways

X X Y X X - 10*9 = 90 ways

X Y Y Y X - 10*9 =90 ways

Total 1000 ways

pnk on October 5th, 2010 at 4:21 am

first place : 10 ways

5th place: 1 ways

2nd place: 10 ways

4th place: 1 way

3rd place: 10 ways

10*1*10*1*10 = 1000 ways

econbddk on October 5th, 2010 at 5:13 am

five letter palindrome can be calculated by 1000 ways, that is, a possible arrangement will be XXCXX OR XXBXX etc. so we have 10 for options for the middle point and we have 10 options for both of the 1st and the 2nd plces from left. As the arrangement is symmetric the options can be calculated as 10*10*10*1*1 =1000 ways.

Ron on October 5th, 2010 at 5:48 am

Answer is 1000 ways.

There are 10C1 (pick 1 letter out of 10 letters) ways of selecting the first letter, second letter and third letter, but there is only 1 way of selecting the fourth and fifth letter as they should match the second and the first letter respectively. Thus, 10C1 x 10C1 x 10C1 x 1C1 x 1C1 = 1000

altud9848 on October 5th, 2010 at 6:57 am

the answer should be 1000 ways because only the first three letters of the arrangements can vary which means we have 10*10*10*1*1=1000 ways to solve.

Gaurav on October 5th, 2010 at 7:36 am

Answer is 1000 Ways.

Lets say the 5-letter word is C1C2C3C2C1 . Each letter C1, C2 and C3 can be chosen in 10 ways.

So, answer is 10*10*10 = 1000 ways.

Anahatha on October 5th, 2010 at 9:23 am

1000

s.sharma on October 5th, 2010 at 9:24 am

the answer is 1000.

we have 5 places. The first place can have any of the 10 alphabets. Therefore, we have 10 ways to fill this place.The last or the fifth place needs exactly the same alphabet as first alphabet. Therefore this place can be filled in just one way.

Similarly, second and the fourth places can be filled in 10 ways and 1 way respectively. For the third place we again have 10 choices.

Therefore by the fundamental principle of multiplication 10 x 10 x 10 x 1 x 1 = 1000.

Yash on October 5th, 2010 at 11:07 am

My answer is 1000.

1st letter - 10 options

2nd letter - 10 options

3rd letter - 10 options

4th letter - 1 option (would be same as 2nd)

5th letter - 1 option (would be same as 1st)

Total options = 10 * 10 * 10 * 1 * 1 = 1000

Eric Bahn on October 5th, 2010 at 8:30 pm

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