karishmatandon wrote:
manishuol wrote:
Each student at a certain university is given a four-character identification code, the Â…rest two characters of which are digits between 0 and 9, inclusive, and the last two characters of which are selected from the 26 letters of the alphabet. If characters may be repeated and the same characters used in a different order constitute a different code, how many different identification codes can be generated following these rules?
A . 135,200
B. 67,600
C. 64,000
D. 60,840
E. 58,500
Four character identification code
_ _ _ _
First two parts for the code, are digits between 0-9, therefore, 10 options for the first part of the code,
and as characters may be repeated, 10 options for the second part as well
Therefore, we have 10 X 10 possibilities for the first and second part of the code
Last two parts of the code, are characters selected from the 26 letters of the alphabet, therefore, 26 options for the third part of the code,
and as characters may be repeated, 26 options for the fourth part as well
Therefore, we have 26 X 26 possibilities for the third and fourth part of the code
so, in all total no. of different identification codes generated following these rules
= 10 X 10 X 26 X 26 = 67600
Answer B
Great explanation of the governing concept. I'd like to add something, though.
Since time is a concern I found myself using this shortcut:
Instead of 10 x 10 x 26 x 26 I used 10 x 10 x 26 x 25.
Changing the last 26 to a 25 allowed me to multiply by 100 then divide by 4 (=25) instead of multiplying by another 26.
10 x 10 x 26 x 100 / 4
10 x 10 x 26 = 2600, 2600 x 100 = 260.000, 260.000 / 4 = 65.000
You know that 26 is barely above 25, so you're looking for the answer that is above, but close to 65.000. 67.600 is the clear answer. Saved myself a good 20-30 seconds because I could do the whole calc in my head.
Another look at it shows me a quicker way, as soon as you know that its 10 x 10 x 26 x 26 you know that the answer's last non-zero digit will be a 6 (you're first non-zero calculation in long multiplication will be 6 x 6 [= 36]). There's only one choice!
Just my $0.02
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