Thread: How to solve this permutation and combination question? Please help

1. How to solve this permutation and combination question? Please help

Find how many numbers greater than 2500 can be formed from the digits 0,1,2,3,4 if no digit can be used more than once.
Please explain in details.
I find it so hard to understand this topic.

2. "Find how many numbers greater than 2500 can be formed from the digits 0,1,2,3,4 if no digit can be used more than once.
Please explain in details.
I find it so hard to understand this topic."

Clearly the only way you can get a larger number than 2500 is to either start with the 3 or the 4, any other starting number would be too small

now consider that you start with the number 3, it doesnt mater what order the other numbers go in the final number will always be bigger than 2500.
so you need to find the number of combinations that you can arrange 0,1,2,4 which is given by n!
so in this case is 4!=24

it is exactly the same if you start with 4

so your final answer will be

2x24=48

hope this helps

3. Originally Posted by syster
"Find how many numbers greater than 2500 can be formed from the digits 0,1,2,3,4 if no digit can be used more than once.
Please explain in details.
I find it so hard to understand this topic."

Clearly the only way you can get a larger number than 2500 is to either start with the 3 or the 4, any other starting number would be too small

now consider that you start with the number 3, it doesnt mater what order the other numbers go in the final number will always be bigger than 2500.
so you need to find the number of combinations that you can arrange 0,1,2,4 which is given by n!
so in this case is 4!=24

it is exactly the same if you start with 4

so your final answer will be

2x24=48

hope this helps
A number starting with 0 can still be greater than 2500. For example, 03124. This is still greater than 2500 isn't it.
I suspect there's a error in the question. This question was given by my lecturer. If the question is looking for number greater than 25000, then i can see the answer is 48.

4. oh yea sorry bwt that

well if we assume there isnt a mistake we can do the same sort of thing, by looking at how many numbers are less than 2500 and taking it away from the total

It will have to start with the zero
Then asume the second number is 1 then the other 3 numbers (2,3,4) can be arranged in 3!=6 different ways

now say the second number is 2, we can clearly see the next 3 numbers can be arranged in any order the resulting number will always be less than 2500 for example 02431<2500
so again there are 3!=6 diffent numbers

with a bit of thought you can see there are no other options, since if the second number was 3 the resulting final number will be to large.

so we now know there are 6+6=12 numbers the can be made from 0,1,2,3,4 that are less than 2500

finaly consider the total number of combinations which is as i said before n! giving in this case 5!=120

so your answer must be 120-12=108

hope this wasnt too confusing

5. Originally Posted by azuki
A number starting with 0 can still be greater than 2500. For example, 03124. This is still greater than 2500 isn't it.
I suspect there's a error in the question. This question was given by my lecturer. If the question is looking for number greater than 25000, then i can see the answer is 48.
For example, 03124.
Well that is one of the 48 that has already been counted, the four digit numbers greater than 2500.
Usually, a five digit number cannot begin with zero.
So there are 4 possible first digits leaving 4 other digits.
TOTAL: $4(4!)+48$ greater than 2500.