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 = 10 and r = 4, we get:
| CR(10,4) = | (10 + 4 - 1)! |
| 4!(10 - 1)! |
| CR(10,4) = | 13! |
| 4!(9)! |
Calculate 13!
13! = 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
13! = 6227020800
Calculate 4!
4! = 4 x 3 x 2 x 1
4! = 24
Calculate 9!
9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
9! = 362880
Plug in factorial values:
| CR(10,4) = | 13! |
| 4!(9)! |
| CR(10,4) = | 6227020800 |
| 24(362880) |
| CR(10,4) = | 6227020800 |
| 8709120 |
CR(10,4) = 715
You have 2 free calculationss remaining
Excel or Google Sheets formula:
Excel or Google Sheets formula:=FACT(10+4-1)/FACT(4)(FACT(10 - 1)What is the Answer?
CR(10,4) = 715
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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