Alphabet palindrome * no. not a palindrome + Alphabet not a palindrome * no. a palindrome.

For alphabet to be a Palindrome, lets take one alphabet for ex: A,any of the 26 letters,A. For 26 alphabets it should be 26(1*26*1) = 26^2

For alphabet not to be a palindrome, lets take one alphabet for ex: A, any of the 26 letters, any of the 25 letters except A.
For 26 Alphabets it should be 26(1*26*25)

For number to be a palindrome, same type of calculation will result in: 10(1*10*1)
For number to be not a palindrome: 10(1*10*9)

total = 26(1*26*1)*10(1*10*9) + 26(1*26*25)*10(1*10*1) = 26^2 * 10^2*9 + 26^2 * 25 * 10^2

Total possible combinations = 26^3 * 10^3

Probability = Total / Total possible combinations

= (26^2 * 10^2*9 + 26^2 * 25 * 10^2)/26^3 * 10^3

= 17/130

OA - C

This would be the probability of having a letter palindrome P(LP) OR the probability of having a number palindrome P(NP) less the probability of having a letter palindrome P(LP) AND the probability of having a number palindrome P(NP):

P(LP)= 26*(1/26 * 26/26 *1/26)= 1/26
P(NP)= 10*(1/10*10/10*1/10)=1/10

P(LP u NP)= 1/26 +1/10 - (1/26*1/10)= 35/260

Now, we still need to substract P(LP n NP), because in P(LP u NP) we are counting the outcome in which both happen at the same time and we are required to calculate either one or the other BUT NOT BOTH, so...

P(LP u NP) - P(LP n NP)= 35/260 - 1/260 = 17/130

a quick answer..
Since the palindromes are only 3 letters or digits, the probability would depend on the third digit/letter

so

1) \(L_1-L_2-L_3-D_1-D_2-D_3\)---------- 26-26-1-10-10-9
when palindrome is with letters, the 3-digit number can have any of remaining 9 digits in units digit, as it has to be different from the first digit \(D_1\)
2) \(L_1-L_2-L_3-D_1-D_2-D_3\)---------- 26-26-25-10-10-1
when palindrome is with digits, the 3-letter formed can have 25 ways in third place, \(L_3\), different from the first digit \(L_1\)
so total ways 9+25=34

while in total ways these number will be 26*10=260

P = \(\frac{34}{260}=\frac{17}{130}\)

C
_________________

CASES Total
\(26*26*26*10*10*10 = {26}^3{10}^3\)
\(T.Letters = {26}^3\)
\(T.Digits = {10}^3\)

CASES Palindrome
ABA: 26*25*1 = 650
AAA: 26*1*1 = 26
010: 10*9*1 = 90
000: 10*1*1 = 10

\(P.Letters = 676 = {26}^2\)
\(P.Digits = 100 = {10}^2\)

CASES Either Palindrome Not Both
\([Palindrome.Letters]*[T.digits-100]: [{26}^2]*[{10}^3-{10}^2]\)
\([T.letters-650]*[Palindrome.Digits]: [{26}^3-{26}^2]*[{10}^2]\)

PROBABILITY: Favorable Cases / Total Cases
\(Favorable:[{26}^2]*[{10}^3-{10}^2]+[{26}^3-{26}^2]*[{10}^2]…{26}^2{10}^2(9+25)\)
\(Probability:{26}^2{10}^2(9+25)/{26}^3{10}^3…34/260…17/130\)

Ans (C)

total no of possibilities 26*26*26*10*10*10

alphabhetic palindrome possibilities 26*26*1*10*10*9

numerical possibilities 26*26*25*10*10*1

Therefore the probability = 26*26*10*10 *(9+25) / 26*26*26*10*10*10

=>17 / 130
Therefore IMO C

The letters make a palindrome when the third letter matches the first. Since when we pick the third letter, only 1 of the 26 possible letters will match the first, in 1/26 of all license plates we have a letter palindrome. Similarly, in 1/10 of all license plates we have a number palindrome. The letters and numbers are independent, so in (1/26)(1/10) = 1/260 of all plates we have both types of palindrome. We counted that overlapping case twice, both when we counted the letter palindromes and when we counted the number palindromes, and we didn't want to count that case at all, so we need to subtract it twice, and the answer is

1/26 + 1/10 - 2/260 = 34/260 = 17/130
_________________

we have Arrangements in which the first 3 Slots are filled with Letters and the second 3 Slots are filled with Digits: 26 Possible Letters and 10 Unique Digits

The Total Possible Outcomes, with no constraints and given that there is no restriction on reusing Letters or Digits is:

L --- L --- L ..........D -- D -- D
26 * 26* 26.....*....10 *10 * 10 =

(26)^3 * (10)^3 = DENOMINATOR


to find the Favorable Outcomes, we can break the Successful Arrangements into 2 Scenarios:


Scenario 1: the Letters are a Palindrome --- and --- the Digits are NOT a Palindrome

Arranging the Letters: We can choose any 1 of the 26 Letters to be repeated in the 1st Place and 3rd Place ---> 26 c 1 = 26

and then the Middle Letter can be any 1 of the 26 Letters as a Sequence such as: A -- A -- A : is still considered a palindrome ---> the Arrangement reads the SAME Forward and Backward. Therefore, there are 26 available options of Letters for the 2nd Place

Total Favorable Arrangements for the Letters in Scenario 1 = (26 c 1) * (26) = (26)^2

AND

The three Digits can NOT be a Palindrome. This means whatever Digit we choose for the 1st Place can NOT appear in the 3rd Place.

Thus: we can choose any 1 of the 10 Digits to appear in the 1st Place.

However, whichever Digit has been chosen for the 1st Place can NOT appear in the 3rd Place. The 3rd Place will have 9 available options.

Finally, it does not matter which Digit appears in the Middle/2nd Place. There are 10 available options, as any Digit can fill this space.

Total Favorable Digit Arrangements under Scenario 1 = (10) * (10) * (9) = (10)^2 * (9)


In Summary: Total Favorable OUTCOMES under Scenario 1 = (26)^2 * (10)^2 * (9)

we found the Total Possible Outcomes (DEN) above = (26)^3 * (10)^3

---we can cancel the Powers of (26)^2 and (10)^2 in the NUM and we are left with----

(9) / (26 * 10) = 9 / 260



Scenario 2: the 3 Letters are NOT a Palindrome --- and ---- the 3 Digits ARE a Palindrome

using Similar Logic as we used above for Scenario 1, we can choose any 1 of the 26 Letters to occupy the 1st Place.

However, since the 3 Letters can not be a Palindrome, whichever Digit is chosen in the 1st Place can NOT appear in the 3rd Place. Thus, there are 25 Available Options of Letters to Fill the 3rd Place.

The 2nd Place/Middle Slot can take any 1 of the 26 Letters

(26) * (26) * (25) = (26)^2 * (25)

AND

the 3 Digits MUST be a Palindrome. Therefore, the 1st Place and 3rd Place must contain the Same Digit so that the 3 Digits read the Same going forward and backward. We can choose this same digit to fill both places by choosing any 1 of the 10 available Digits ----> (10 c 3) = 10

the 2nd Place can take any Digit. Thus, there will be 10 available options of Digits to choose from.

(10) * (10) = (10)^2


the Number of Favorable Arrangements under Scenario 2 = (26)^2 * (25) * (10)^2

the Total Possible Arrangements (DEN) we found above as = (26)^3 * (10)^3


---again, we can cancel out the (26)^2 and the (10)^2 in the NUM with the DEN and we are left with


(25) / (26 * 10) = 25/260


Finally, Adding the: (Probability under Scenario 1) + (Probability under Scenario 2) =

(9 / 260) + (25 / 260) = 34/260 = 17/130


*Answer*

17/130