subhro jana

 Profile Answers by 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.)

iluilu

 Profile Answers by iluilu 

• Jun 17th, 2008

 

      20 numbers can be formed

sandi.trivedi

 Profile Answers by sandi.trivedi 

• Jul 7th, 2008

 

twelve

santhosh_cbit

 Profile Answers by santhosh_cbit 

• Jul 20th, 2008

 

1298

m_manic

 Profile Answers by m_manic Questions by m_manic

• Dec 9th, 2009

 

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.

shankarleelap

 Profile Answers by shankarleelap 

• Jan 29th, 2011

 

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