There are 6 digits in the set, giving
A choice of 6 for the first digit, leaving
a choice of 5 for the second digit for each choice of the first, leaving
a choice of 4 for the third digit for each choice of the first two, leaving
a choice of 3 for the fourth digit for each choice of the first three, leaving
a choice of 2 for the fifth digit for each choice of the first four
making:
number of 5-digit numbers = 6 x 5 x 4 x 3 x 2 = 720 possible 5-digit numbers.
This is a choice where the order matters and is called a permutation. To calculate a permutation of r items from a set of n the formula:
nPr = n!/(n - r)!
will do it, where n! means "n factorial" and is calculated as the multiple of all digits less than n but greater than 0, ie:
n! = n x (n - 1) x (n - 2) x ... x 2 x 1
for example, 4! = 4 x 3 x 2 x 1 = 24. (0! - zero factorial - is defined to be 1, ie 0! = 1.)
In this question, n = 6 (set of 6 digits) and a permutation of 5 digits (for a 5-digit number) is required, thus the number of permutations (5-digit numbers) is:
6P5 = 6!/(6 - 5)!
= 6!/1!
= 6 x 5 x 4 x 3 x 2 x 1/1
= 720
If the question had been how many 4-digit numbers could be made from that set of 6 digits, then the answer would have been:
6P4 = 6!/(6 - 2)!
= 6 x 5 x 4 x 3 x 2 x 1/2 x 1
= 6 x 5 x 4 x 3
= 360
1,956 different numbers can be made from 6 digits. You can calculate this by using the permutation function in a summation function, like this: Σ6k=1 6Pk = 6P1+6P2+...+6P5+6P6 What this does is calculate how many 1 digit numbers you can make from 6 digits, then how many 2 digit numbers can be made from 6 digits and adds the amounts together, then calculates how many 3 digit numbers can be made and adds that on as well etc.
The answer is 10C4 = 10!/[4!*6!] = 210
210
5040 numbers can be made.
-- If the same digit may be repeated, then 64 can be made. -- If the same digit may not be repeated, then 24 can be made.
120 5-digit numbers can be made with the numbers 12345.
1,956 different numbers can be made from 6 digits. You can calculate this by using the permutation function in a summation function, like this: Σ6k=1 6Pk = 6P1+6P2+...+6P5+6P6 What this does is calculate how many 1 digit numbers you can make from 6 digits, then how many 2 digit numbers can be made from 6 digits and adds the amounts together, then calculates how many 3 digit numbers can be made and adds that on as well etc.
0 you can only have 3 layers since you have only 3 numbers so without repeating you would only have 3 layers
a lot
The answer is 10C4 = 10!/[4!*6!] = 210
2
210
9
You can make 4*3*2/2 = 12 numbers.
5040 numbers can be made.
The general formula is to square the number of digits it could be (in this case 5) by how many spaces there are. So in this case it would be 53 = 125 possible combinations.
-- If the same digit may be repeated, then 64 can be made. -- If the same digit may not be repeated, then 24 can be made.