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 = 11 and r = 5, we get:
| CR(11,5) = | (11 + 5 - 1)! |
| 5!(11 - 1)! |
| CR(11,5) = | 15! |
| 5!(10)! |
Calculate 15!
15! = 15 x 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
15! = 1307674368000
Calculate 5!
5! = 5 x 4 x 3 x 2 x 1
5! = 120
Calculate 10!
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
10! = 3628800
Plug in factorial values:
| CR(11,5) = | 15! |
| 5!(10)! |
| CR(11,5) = | 1307674368000 |
| 120(3628800) |
| CR(11,5) = | 1307674368000 |
| 435456000 |
CR(11,5) = 3003
You have 2 free calculationss remaining
Excel or Google Sheets formula:
Excel or Google Sheets formula:=FACT(11+5-1)/FACT(5)(FACT(11 - 1)What is the Answer?
CR(11,5) = 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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