How many combinations can we have from:
r elements from n distinct objects
where order does matter and
replacements are not allowed?
Combinations with Replacment Formula
| CR(n,r) = | (n + r - 1)! |
| r! (n - 1)! |
Plug in n = 9 and r = 6, we get:
| CR(9,6) = | (9 + 6 - 1)! |
| 6!(9 - 1)! |
| CR(9,6) = | 14! |
| 6!(8)! |
Calculate 14!
14! = 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
14! = 87178291200
Calculate 6!
6! = 6 x 5 x 4 x 3 x 2 x 1
6! = 720
Calculate 8!
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
8! = 40320
Plug in factorial values:
| CR(9,6) = | 14! |
| 6!(8)! |
| CR(9,6) = | 87178291200 |
| 720(40320) |
| CR(9,6) = | 87178291200 |
| 29030400 |
CR(9,6) = 3003
You have 2 free calculationss remaining
Excel or Google Sheets formula:
Excel or Google Sheets formula:=FACT(9+6-1)/FACT(6)(FACT(9 - 1)What is the Answer?
CR(9,6) = 3003
How does the Combinations with Replacement Calculator work?
Free Combinations with Replacement Calculator - Calculates the following:
How many combinations can we have from a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed?
This calculator has 2 inputs.
How many combinations can we have from a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed?
This calculator has 2 inputs.
What 1 formula is used for the Combinations with Replacement Calculator?
What 3 concepts are covered in the Combinations with Replacement Calculator?
- combination
- a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter
nPr = n!/r!(n - r)! - combinations with replacement
- factorial
- The product of an integer and all the integers below it
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