# permutation problemWatch

#1
hey

so the question is:
there 26 letters in the alphabet and 10 numerical digits. A car has a registration number consisting of 3 letters followed by 2 digits.

a)How many possiblities are there? 26(26)(26)(10)(10) = 1757600

b)How many possiblities if all the letters must be different? 26(25)(24)(10)(10) = 1560000

c)If there is a free choice of letters, but the digits cannot begin with a 0, and must form an even number, how many possibilities are there? 26(26)(26)(9)(4) = 632736

0
7 years ago
#2
There are 5 even digits (not 4), and it doesn't say the second digit can't be 0.
#3
oh i forgot the zero.. lol thanks!
0
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