How many distinct car numbers can be formed using two letters and three digits? The two letters must be distinct and first digit should be non zero?

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subhro jana  

  • Member Since Feb-2008 | Feb 12th, 2008


the two letters can be chosen in 26p2 ways

the first digit can be chosen in 9 ways(excepting 0)
2nd digit in 10 ways
3rd digit also in 10 ways.
so the three no.s can be chosen in 900(9*10*10).

along with the letters it can be chossen in 26p2*900 ways= 585000ways(plz. check calc.)

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subhro jana

  • Feb 12th, 2008
 

the two letters can be chosen in 26p2 ways

the first digit can be chosen in 9 ways(excepting 0)
2nd digit in 10 ways
3rd digit also in 10 ways.
so the three no.s can be chosen in 900(9*10*10).

along with the letters it can be chossen in 26p2*900 ways= 585000ways(plz. check calc.)

First letter can be selected in 26 ways. Second can be in 25 ways (Two letters must be distinct). Then first digit in 9 ways then second and third in 10 ways. Total number of ways is 585000.

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If the car number is of size 5, then here we have two cases one is the car no
starting with digit and other is no starting with character
Case 1: first place can be occupied in 9 places and remaining 4 places contain 2
characters which can be selected in c(26,2) ways. and other 2 digits can be
selected in 10*10 ways. therefore totally remaining 4 places can be filled in
c(26,2)*10*10*4!.
=> completely 5 places in 9*c(26,2)*10*10*4!
case 2: first place can be occupied by character in 26 ways
and remaining 1 character can be obtained in 25 ways
and remaining 3 digits can be obtained in 10*10*10 ways
thus these 4 places can be arranged in 4! ways
=> completely 5 places in 26*25*10*10*10*4!


Combining 2 cases we have (9*c(26,2)*10*10*4!)+(26*25*10*10*10*4!) ways

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