Bunuel wrote:

A license plate number is created by writing a sequence of three letters followed by a sequence of 3 digits. Any digit 0 - 9 is acceptable, as is any letter. What is the probability that either (but not both) the letters or numbers form a palindrome, a series of letters or numbers that read the same forwards and backwards? (For example, ADA and 121 are palindromes but ABC and 123 are not. ADA121 would not be valid but ADA123 would be valid.)

A. 1/26

B. 1/10

C. 17/130

D. 7/52

E. 9/65

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

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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