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?
468000 b] 676000 c] 608400 d] 585000
for ex ap 539
Since there are 26 letters and the other one is different you are using for the second on 25. Now the first digit cannot be zero thus there are 9 possibilities. In the final two digits there are no restrictions thus there are 10. In total you then have,Originally Posted by bharathlece
$\displaystyle 26(25)(9)(10)(10)=58500$
Greetings:
PHkr's solution would be correct if the two letters and three digits were required to occupy specific positions within the five character succession, e.g., letters followed by digits, letter in first and last position with digits between. Because no such rule of order is specified, we must consider, for instance, that the letters can assume any of 5c2 different configurations (e.g., positions 1 and 2, 1 and 3, etc.). Hence the solution put forth by PHkr need only be modified by a factor of 5c2=10, putting the number of arrangements at 10(26)(25)(9)(10)(10).
Regards,
Rich B.
[QUOTE=Rich B.]Greetings:
PHkr's solution would be correct if the two letters and three digits were required to occupy specific positions within the five character succession, e.g., letters followed by digits, letter in first and last position with digits between. Because no such rule of order is specified, we must consider, for instance, that the letters can assume any of 5c2 different configurations (e.g., positions 1 and 2, 1 and 3, etc.). Hence the solution put forth by PHkr need only be modified by a factor of 5c2=10, putting the number of arrangements at 10(26)(25)(9)(10)(10).
Which is nice, in that the proposed solution is now one of the
options offered.
RonL