security codes can be made with 2 digits followed by 3 letters or 3 digits followed by 2 letters, the digits can contain numbers from 0 to 9 and letters from A to Z but cannot be used twice. how many codes are possible if a code must contain a 9...I know that it's (1*9*26*25*24)+(1*9*8*26*25) but why do you then multiply the first bracket by factorial 2 but the second only by 3?

Question

anonymous · Answer

Because of repeat :)

anonymous · Answer

i don't understand still, sorry

anonymous · Answer

where is the repeat? because with the three digits can't there still be 3! different ways?

anonymous · Answer

Write the ways and find it :)

anonymous · Answer

The 2 and 3 represent how many places you can stick the 9.

anonymous · Answer

Shall I explain it?

anonymous · Answer

yes i understand that but why is it only 3 and not 3! since isn't there also 3! ways? how do you figure our the repeats and then what to multiply it by without writing out all the possibilities?

agent0smith · Answer

If you can put the 9 either here ===>  _, here ===>  _, or here ===> _, how many places are there to put it? 3, or 3! (which is 6)?

anonymous · Answer

but the answer says only multiplied by 3 not 3!, but the first bracket is multiplied by 2!???

agent0smith · Answer

What is 2! equal to?

agent0smith · Answer

I have no idea why they'd write 2! when it's not necessary, though.

agent0smith · Answer

But, again... If you can put the 9 either here ===>  _, here ===>  _, or here ===> _, how many places are there to put it? 3, or 3! (which is 6)?

anonymous · Answer

so it should be 3!?

anonymous · Answer

how do you know when to multiply by a factorial since for another similar question:a security code consists of 3 different letters (A-Z) and number 1-9 and for the first one with what is the probability that it does not contain an A there is no multiplier, but for the second with one that does contain an A there is a multiplier by 3?

agent0smith · Answer

If you can put the 9 either here ===>  _, here ===>  _, or here ===> _, how many places are there to put it? Hint: count how many places there are.

anonymous · Answer

i know there's 3 but within those 3 it can be placed in 3! different ways...

agent0smith · Answer

That doesn't make sense. There are 3 places it can go, there is not 6 ways it can be put into those 3 places. It can go in the first, second, or third place.

You multiply by a factorial when there's a GROUP of objects you can arrange, not when there's a single object you can put into a number of places.

anonymous · Answer

then why is there a factorial with the first (of the second question) and not the second?

anonymous · Answer

shouldn't it be multiplied by 3 in both cases?

anonymous · Answer

with and without the A?

agent0smith · Answer

It'd be easier if you wrote out the question more clearly. I didn't see any mention of a factorial multiplier in the second question 

"for the first one with what is the probability that it does not contain an A there is no multiplier, but for the second with one that does contain an A there is a multiplier by 3?"

agent0smith · Answer

You don't need a multiplier of 3 for this: 

a security code consists of 3 different letters (A-Z) and number 1-9... does not contain an A 

you can't use an A, so there's 25 letters to choose from at the start. Number of codes not containing an A:

25*24*23*9


number of codes containing an A:
(1*25*24*9)*3 <=== since there's 3 places we can put the A.

anonymous · Answer

understood

anonymous · Answer

what would be an instance to mulitply by a factorial?

agent0smith · Answer

When there's objects or groups of objects that you can arrange/put in order. I'm too tired to think of a good example right now.