- #1
cronuscronus
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Hi all. I wanted to double check some of my work, and get some feedback if there's any errors. Any help is appreciated!
a) A particular digit (say 5, for example).
Essentially we're choosing one digit. 10C1 is 10, and we have 5 slots, so the answer here is 10^5 = 100,000 combinations.
b) A pair and three other distinct digits. (e.g 27421)
Here we are really just choosing 4 distinct digits. So we calculate the combinations without replacement. 10*9*8*7 = 5,040 combinations.
c) Two pairs of distinct digits and 1 other distinct digit. (e.g
12215)
Here we are choosing 3 distinct digits. So we calculate the combinations without replacement. 10*9*8 = 720 combinations.
d) A palindrome of length 5 (e.g. 12321 , or 11111 or 12221)
The first digit can be selected from 1-9. The next digit can be selected from 0-9. The next digit can also be selected from 0-9. The last two digits are determined.
So we can calculate 9*10^2 = 900 5-digit palindromes.
Thanks for the help!
a) A particular digit (say 5, for example).
Essentially we're choosing one digit. 10C1 is 10, and we have 5 slots, so the answer here is 10^5 = 100,000 combinations.
b) A pair and three other distinct digits. (e.g 27421)
Here we are really just choosing 4 distinct digits. So we calculate the combinations without replacement. 10*9*8*7 = 5,040 combinations.
c) Two pairs of distinct digits and 1 other distinct digit. (e.g
12215)
Here we are choosing 3 distinct digits. So we calculate the combinations without replacement. 10*9*8 = 720 combinations.
d) A palindrome of length 5 (e.g. 12321 , or 11111 or 12221)
The first digit can be selected from 1-9. The next digit can be selected from 0-9. The next digit can also be selected from 0-9. The last two digits are determined.
So we can calculate 9*10^2 = 900 5-digit palindromes.
Thanks for the help!