sanju09 wrote:How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?
(A) 1875
(B) 2000
(C) 2375
(D) 2500
(E) 3875
[spoiler]Source: Eric's collection on BTG
OA C[/spoiler]
Here following cases are possible:-
1) when 4 is present at the leftmost digit.
2) when 4 is present at the digit adjacent to the left most digit,
3) when 4 is not present at any of the leftmost places;
case 1) -,-,-,-,-; when 4 is present at the left most digit, then the digit adjacent to it can be filled (by any of the (0,2,6,8)) in 4 ways, and the remaining places can be filled by the 5 odd numbers in 5*5*5 ways; hence total no. of ways= 1*4*5*5*5;
case 2)-,-,-,-,-; when 4 is not there at the left most digit, it can be filled ( by any of the 2,6,8) in 3 ways, and the digit adjacent to the leftmost digit will be filled by the 4 in 1 way; and the remaining places can be filled by the 5 odd numbers in 5*5*5 ways; hence total no. of ways= 3*1*5*5*5;
case 3)-,-,-,-,-; when 4 is not selected for the arrangement; then the left most digit can be filled ( by any of the 2,6,8) in 3 ways, and the digit adjacent to leftmost digit can be filled ( by any of the 0,2,6,8) in 4 ways, and the remaining places can be filled by the 5 odd numbers in 5*5*5 ways; hence total no. of ways = 3*4*5*5*5;
hence total required no. of ways= 1*4*5*5*5+3*1*5*5*5+3*4*5*5*5=2375 hence
CO Excellence... my search for you is on... you can be far.. but not beyond my reach!
