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# Difficult to phrase question to do with number of possibilities Watch

1. If there were 4 destinations and you had to visit each one in a single journey. What is the number of possible paths you could take to reach them all?

Would the calculation be 44 ?
2. (Original post by BhagwanNoBhool)
If there were 4 destinations and you had to visit each one in a single journey. What is the number of possible paths you could take to reach them all?

Would the calculation be 44 ?
Could you provide a picture of the question?
3. (Original post by BhagwanNoBhool)
If there were 4 destinations and you had to visit each one in a single journey. What is the number of possible paths you could take to reach them all?

Would the calculation be 44 ?
Call the destinations A, B, C, D

then a possible path could be BCAD or ADBC

Consider how many options ther are for the first letter... then the second letter etc.
4. (Original post by Zacken)
Could you provide a picture of the question?
It's not a written question, but maths problem to do with a computer program for my course work

(Original post by notnek)
Call the destinations A, B, C, D

then a possible path could be BCAD or ADBC

Consider how many options ther are for the first letter... then the second letter etc.
Would there be 42 for each letter? So 64 paths in total?
5. (Original post by BhagwanNoBhool)
Would there be 42 for each letter? So 64 paths in total?
You seem to be guessing or using some incorrect method without thinking.

ACBD
DACB
etc.

How can there be 16 choices for the first letter?
6. (Original post by notnek)
You seem to be guessing or using some incorrect method without thinking.

ACBD
DACB
etc.

How can there be 16 choices for the first letter?
Sorry, I misinterpreted what you were saying.
There are 4 choices for each letter.

So 4 paths for the first one,
A
B
C
D

Then 4 times as many for the next.
AA CA
AB CB
AC CC
BA DA
BB DB
BC DC
BD DD

Would it always be four times as many for the next letter?
7. (Original post by BhagwanNoBhool)
Sorry, I misinterpreted what you were saying.
There are 4 choices for each letter.

So 4 paths for the first one,
A
B
C
D
That's correct.

To visit all four destinations you need the four letters to be different e.g. ACDB or CDBA. That means the second letter cannot be the same as the first letter.

So if there are 4 choices for the first letter, how many choices are there for the second?

This is all assuming I'm interpreting your question correctly. Some of your replies suggest that your actual question may be different.
8. (Original post by notnek)
That's correct.

To visit all four destinations you need the four letters to be different e.g. ACDB or CDBA. That means the second letter cannot be the same as the first letter.

So if there are 4 choices for the first letter, how many choices are there for the second?

This is all assuming I'm interpreting your question correctly. Some of your replies suggest that your actual question may be different.
No you're correct with the question, sorry I wasn't thinking with the repeating letter ones. For the second there are 3 choices remaining, 2 for the third and 1 for the last.
9. (Original post by BhagwanNoBhool)
No you're correct with the question, sorry I wasn't thinking with the repeating letter ones. For the second there are 3 choices remaining, 2 for the third and 1 for the last.
That's correct. Multiply those numbers together to find the total possibilities.
10. (Original post by notnek)
That's correct. Multiply those numbers together to find the total possibilities.
4*3*2*1 = 24

So 24 different paths?
11. (Original post by BhagwanNoBhool)
4*3*2*1 = 24

So 24 different paths?
Correct
12. (Original post by notnek)
Correct
Thanks man
It seems really obvious in hindsight, but I wasn't thinking about it properly

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