How many 7-number license plates are possible? SS Provided below~
The way they explain it in your screenshot is confusing haha But here's the concept We have a 7 number license plate |dw:1614180331581:dw|
We can have a license plate number of 0000000 or 1234567 and so many more different possibilities We're trying to calculate all those possibilities
Right
For each blank, we have 10 choices We can have either 0, 1, 2, 3, 4, 5, 6 , 7, 8, 9 Following along still?
Yes c:
so |dw:1614180684465:dw|
we can have 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 for the first digit 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 for the second digit ... and so on until 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 for the 7th digit
is it 10 to the power of 7? or 7 to the power of 10 lol
so what are the total possibilities going to be if we have 7 digits and 10 choices for every digit
\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531 is it 10 to the power of 7? or 7 to the power of 10 lol \(\color{#0cbb34}{\text{End of Quote}}\) I'm not really sure, lol
The slide in your screenshot says if we make 'r' selections from 'n' choices total number of arrangements is \(\Large n^r\) Another way to say selections is digits and choices is choices
and I'm not really sure ;/
Ok Im going to make this a little easier how many times does 7 go into 10 and how many times does 10 go into 7?
70, but thats not the answer-
To find the total number of possibilities, you just multiply like this |dw:1614180919388:dw|
so if you're multiply 10 by itself seven times is that 10^7 or 7^10
10 ^7
That's your answer! Were you able to understand though? For each blank, we could choose from any of the 10 numbers. Since we have 7 blanks, and each one can be one of those 10 numbers. Then the total number of possibilities is 10*10*10*10*10*10*10 or 10^7 like you said
Hmmm, well I googled the question, because I was confused, and well one answer said: There are 26 letters and 10 digits, or 36 possible characters. Assuming all combinations are allowed, this is 36^7 = 78,364,164,096 possible combinations.
Also AZ is right but a good thing to do is always double check your work to make sure it right.
its*
Okai, I'll try it, ty <3
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