No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once Solution

STEP 0: Pre-Calculation Summary
Formula Used
Number of Combinations = (Value of P+1)*(Value of Q+1)*(2^Value of N)-1
C = (p+1)*(q+1)*(2^n)-1
This formula uses 4 Variables
Variables Used
Number of Combinations - Number of Combinations is defined as the total number of unique arrangements that can be made from a set of items, without regard to the order of the items.
Value of P - Value of P is any natural number or positive integer that can be used for combinatorial calculations.
Value of Q - Value of Q is any natural number or positive integer that can be used for combinatorial calculations.
Value of N - Value of N is any natural number or positive integer that can be used for combinatorial calculations.
STEP 1: Convert Input(s) to Base Unit
Value of P: 7 --> No Conversion Required
Value of Q: 6 --> No Conversion Required
Value of N: 8 --> No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
C = (p+1)*(q+1)*(2^n)-1 --> (7+1)*(6+1)*(2^8)-1
Evaluating ... ...
C = 14335
STEP 3: Convert Result to Output's Unit
14335 --> No Conversion Required
FINAL ANSWER
14335 <-- Number of Combinations
(Calculation completed in 00.004 seconds)

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Combinations Calculators

No of Combinations of N Different Things taken R at once given M Specific Things Always Occur
​ LaTeX ​ Go Number of Combinations = C((Value of N-Value of M),(Value of R-Value of M))
No of Combinations of N Different Things taken R at once and Repetition Allowed
​ LaTeX ​ Go Number of Combinations = C((Value of N+Value of R-1),Value of R)
No of Combinations of N Different Things taken R at once given M Specific Things Never Occur
​ LaTeX ​ Go Number of Combinations = C((Value of N-Value of M),Value of R)
No of Combinations of N Different Things taken R at once
​ LaTeX ​ Go Number of Combinations = C(Value of N,Value of R)

No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once Formula

​LaTeX ​Go
Number of Combinations = (Value of P+1)*(Value of Q+1)*(2^Value of N)-1
C = (p+1)*(q+1)*(2^n)-1

What are Combinations?

In combinatorics, Combinations refer to the different ways of selecting a subset of items from a larger set without regard to the order of selection. Combinations are used to count the number of possible outcomes when the order of selection does not matter. For example, if you have a set of three elements {A, B, C}, the Combinations of size 2 would be {AB, AC, BC}. In this case, the order of the items within each combination does not matter, so {AB} and {BA} are considered the same combination.

The number of Combinations of selecting "k" items from a set of "n" items is denoted as C(n, k). It is calculated using the binomial coefficient formula: C(n, k) = n! / (k! * (n - k)!)

Combinations have various applications in mathematics, probability theory, statistics, and other fields.

How to Calculate No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once?

No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once calculator uses Number of Combinations = (Value of P+1)*(Value of Q+1)*(2^Value of N)-1 to calculate the Number of Combinations, The No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once formula is defined as the total number of ways of selecting one or more things out of (p+q+n) things, where ‘p’ identical things of one type ‘q’ identical things of another type and ‘n’ different things. Number of Combinations is denoted by C symbol.

How to calculate No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once using this online calculator? To use this online calculator for No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once, enter Value of P (p), Value of Q (q) & Value of N (n) and hit the calculate button. Here is how the No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once calculation can be explained with given input values -> 7167 = (7+1)*(6+1)*(2^8)-1.

FAQ

What is No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once?
The No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once formula is defined as the total number of ways of selecting one or more things out of (p+q+n) things, where ‘p’ identical things of one type ‘q’ identical things of another type and ‘n’ different things and is represented as C = (p+1)*(q+1)*(2^n)-1 or Number of Combinations = (Value of P+1)*(Value of Q+1)*(2^Value of N)-1. Value of P is any natural number or positive integer that can be used for combinatorial calculations, Value of Q is any natural number or positive integer that can be used for combinatorial calculations & Value of N is any natural number or positive integer that can be used for combinatorial calculations.
How to calculate No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once?
The No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once formula is defined as the total number of ways of selecting one or more things out of (p+q+n) things, where ‘p’ identical things of one type ‘q’ identical things of another type and ‘n’ different things is calculated using Number of Combinations = (Value of P+1)*(Value of Q+1)*(2^Value of N)-1. To calculate No of Combinations of N Different Things, P and Q Identical Things taken Atleast One at once, you need Value of P (p), Value of Q (q) & Value of N (n). With our tool, you need to enter the respective value for Value of P, Value of Q & Value of N and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Number of Combinations?
In this formula, Number of Combinations uses Value of P, Value of Q & Value of N. We can use 3 other way(s) to calculate the same, which is/are as follows -
  • Number of Combinations = C(Value of N,Value of R)
  • Number of Combinations = C((Value of N+Value of R-1),Value of R)
  • Number of Combinations = C((Value of N-Value of M),(Value of R-Value of M))
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