Permutations

[Script Coder]

I have been doing my AP Math homework on permutations and I have realised that my answers differ a lot from the answers I am supposed to get. Now I realise that I could very well be wrong, but I just want to know if any of you would mind helping me out. The problems are:

1) How many numbers, divisible by 5, can be formed from the digits 3,6,7,5. Each digit may only be used once in each number?
My answer: 6
Book's answer: 16
2) How many numbers can be formed from 0,3,4,5,6,7 given that each number may only be used once, the number must be strictly larger than 50000 and it must be even?
My answer:192
Book's answer: 504

3) This last one I get the answer, but only by trying all permutations, is there a faster way:
How many words can be formed by choosing three letters from the word "PIPPIN"?
My answer: 19
Book's answer: 19

EDIT: Making it more readable

[giblit]

1) I think the trick part of this question is it doesn't say how many digits the number must be. If it had to be 4 digits it would be 6. But there is no limit.

the last two digits must always be 35 , 65, or 75
For 1 digit there is 1 choice: 5
for 2 digits there is 35 , 65 , 75 = 3 choices
3 digits: 365 , 375 , 635 , 735 , 675 , 765 = 6 choices
4 digits: 3675 , 3765 , 7635 , 7365 , 6735 , 6375 = 6 choices
1 + 3 + 6 + 6 = 4 + 12 = 16



2) Basically to be larger than 50,000 you must have either a 5 digit number that starts with 5 , 6 , or 7 or a 6 digit number that starts with 3 , 4 , 5 , 6 , or 7. Then to be even it will have to end in 0 , 4 or 6.

So lets look at this.

5 digit number:

for the first and last digit we can do

50 , 54 , 56
60 , 64
70 , 74 , 76
which is a possible of 8 choices.
digits 2-4 have 4 * 3 * 2 * 1 = 24 choices

So all together we have 8 * 24 = 192 choices


6 digit number:
For the first and last digit the possible choices are:
30 , 34 , 36
40 , 46
50 , 54 , 56
60 , 64
70 , 74 , 76
which is a possible of 13 choices

digits 2-5 have 4 * 3 * 2 * 1 = 24 choices

so all together we have 13 * 24 = 312 choices


So for a 5 or 6 digit number we have 192 + 312 choices = 504 choices.


3) The possible permutations with out repetition is 6.
Which are:
PIN , PNI , INP , IPN , NIP , NPI

Then with repetition
We have
PPP , IPP , PIP , PPI , NPP , PNP , NPP
IIP , IPI , PII , IIN , INI , NII

So now we have 6 + 7 + 6 = 19

Basically there are 3 letters to pick from p, i , n
So the non-repetition would be 3! = 6

Then there are 3 P's , and 2 I's

Then there is 1 option to use all 3 's.

Then when we use 2 of the same letters it can go 3 different ways.
xYY , YYx , YxY There are then 2 other letters to choose from so this gives us 3 * 2 for 1 duplicate but we have 2 duplicates so this will be 3 * 2 * 2 = 12

so we have 6 + 1 + 12 = 19

[Script Coder]

Thank you so much, you are a life saver. I see now that they really enjoy tricking you.

[ResidentBiscuit]

That's the main thing with permutations is just figuring exactly what is being asked. Once you have that down, they're mostly pretty simple.