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Ok my iPhone have a free apple calc but I have to pay for one on my iPad? Or be harassed once this monstrosity loads with non stop analytical ads with geolocates you guys put a whole lotta thought into making things way more difficult than they have to be.

I'm inconvenienced to say the least I hope iPhone x craps out, and apple shares plummet lol.

I like my iPad, not impressed with apple right now tho.

Ads are so prevalent, it's impossible to use

This is an example of a very aggressive model for ads that ruins the user experience. Yes, I could purchase the ad-free version, but within 5 minutes I was so annoyed that I removed the application. I can appreciate that the developers need to make money and I would gladly pay if they weren't so outrageously aggressive in forcibly placing ads that block calculations on screen.


Our apologies for the advertisements in the ads. We are constantly working hard on maintaining and improving our app, so we can offer the best product to our users for free. Unfortunately we have to use advertisement in order to fund this. We do want to thank you for your feedback, we will use this to balance the amount of ads better in the future.

I would also like to point out that our ads should not block out any calculations on screen, so if this happened then it's most likely a bug. We would like to fix this as soon as possible, and could use your help in this. If could send us a screenshot of this happening to [email protected] that would be greatly appreciated.


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Erlang Calculator - for Call Centre Staffing (Online Version )

Now with Week and Month Planner and Upload Facility

Statistics from All Erlang Calculations

Number of Erlang Calculations performed
Past 24 HoursPast Month

Average Values Entered
AHT (Seconds)
AHT (Minutes)
Average Service Level%
Average Target Answer Time (Seconds)
Average Shrinkage27%
Average Max Occupancy%
Above figures include calls and other work tasks

How To Use This Erlang Calculator

  1. If you have calls per hour, then enter the number of incoming contacts as and the period is 60 minutes.
  2. The Average Handling Time is the amount of time that a person (an agent) takes to handle a phone contact. This includes the talk time as well any paperwork time (wrap-up time) before they are able to answer the next call. This should be in seconds.
  3. Put in your Service Level target and time. So if you wanted to handle 90% of calls in 15 seconds, put in 90 and If you are uncertain of this the industry "average" is 80% of calls answered in 20 seconds.
  4. This contact centre staffing Erlang calculator is a hybrid model based on both the Erlang C formula the Erlang A formula. The Erlang C formula was invented by the Danish Mathematician A.K. Erlang and is used to calculate the number of advisors and the service level. Call Abandons are calculated using the Erlang A formula which was devised by Swedish statistician Conny Palm in This assumes an Average Patience - also know as Average Time to Abandon (ATA).
  5. We also have a more flexible Microsoft Excel based version of this calculator. You can download the free Excel Erlang Calculator
  6. This calculator works on probabilities, so may appear to overstate the number of agents needed at low levels. So for example if you enter 0 calls per hour it will say that you need 1 agent. This is quite correct, as there may be a possibility that one call may come in. In practicality, you may decide to not schedule any staff.
  7. The maximum number of agents that the calculator can calculate before shrinkage is applied is 10, Agents.
  8. Shrinkage is a factor designed to take into account holidays, sickness etc. For more information Read this article on how to calculate shrinkage
  9. The maximum occupancy is designed to improve accuracy. If you take Occupancy over 85% - 90% for long periods you will find that it gets hidden in a longer AHT figure, and agent burn out happens.
  10. Call Abandons are calculated using the Erlang A formula, which assumes an Average Patience -also know as Average Time to Abandon (ATA).
  11. The Calculator can deal with up to 10, agents, thanks to some help with the maths from Philip Wright CEng – (Former Technical Director & CTO Europe at Aspect Telecommunications/Communications ).

Need To Include WebChat And Emails?

Need a Multi-Channel calculator? Then use our free Multi Channel Calculator

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Building a Basic Calculator - C - Tutorial 13

Combinations Calculator (nCr)

Calculator Use

The Combinations Calculator will find the number of possible combinations that can be obtained by taking a sample of items from a larger set. Basically, it shows how many different possible subsets can be made from the larger set. For this calculator, the order of the items chosen in the subset does not matter.

There are n! ways of arranging n distinct objects into an ordered sequence, permutations where n = r.
The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are not allowed.
The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed.  When n = r this reduces to n!, a simple factorial of n.
Combination Replacement
The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are allowed.
Permutation Replacement
The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are allowed.
the set or population
subset of n or sample set


Combinations Formula:

\( C(n,r) = \dfrac{n!}{( r! (n - r)! )} \)

For n ≥ r ≥ 0.

The formula show us the number of ways a sample of “r” elements can be obtained from a larger set of “n” distinguishable objects where order does not matter and repetitions are not allowed. [1] "The number of ways of picking r unordered outcomes from n possibilities." [2]

Also referred to as r-combination or "n choose r" or the binomial coefficient.  In some resources the notation uses k instead of r so you may see these referred to as k-combination or "n choose k."

Combination Problem 1

Choose 2 Prizes from a Set of 6 Prizes

You have won first place in a contest and are allowed to choose 2 prizes from a table that has 6 prizes numbered 1 through 6. How many different combinations of 2 prizes could you possibly choose?

In this example, we are taking a subset of 2 prizes (r) from a larger set of 6 prizes (n). Looking at the formula, we must calculate “6 choose 2.”

C (6,2)= 6!/(2! * ()!) = 6!/(2! * 4!) = 15 Possible Prize Combinations

The 15 potential combinations are {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}, {4,5}, {4,6}, {5,6}

Combination Problem 2

Choose 3 Students from a Class of 25

A teacher is going to choose 3 students from her class to compete in the spelling bee. She wants to figure out how many unique teams of 3 can be created from her class of

In this example, we are taking a subset of 3 students (r) from a larger set of 25 students (n). Looking at the formula, we must calculate “25 choose 3.”

C (25,3)= 25!/(3! * ()!)= 2, Possible Teams

Combination Problem 3

Choose 4 Menu Items from a Menu of 18 Items

A restaurant asks some of its frequent customers to choose their favorite 4 items on the menu. If the menu has 18 items to choose from, how many different answers could the customers give?

Here we take a 4 item subset (r) from the larger 18 item menu (n). Therefore, we must simply find “18 choose 4.”

C (18,4)= 18!/(4! * ()!)= 3, Possible Answers

Handshake Problem

In a group of n people, how many different handshakes are possible?

First, let's find the total handshakes that are possible. That is to say, if each person shook hands once with every other person in the group, what is the total number of handshakes that occur?

A way of considering this is that each person in the group will make a total of n-1 handshakes. Since there are n people, there would be n times (n-1) total handshakes. In other words, the total number of people multiplied by the number of handshakes that each can make will be the total handshakes. A group of 3 would make a total of 3() = 3 * 2 = 6. Each person registers 2 handshakes with the other 2 people in the group; 3 * 2.

Total Handshakes = n(n-1)

However, this includes each handshake twice (1 with 2, 2 with 1, 1 with 3, 3 with 1, 2 with 3 and 3 with 2) and since the orginal question wants to know how many different handshakes are possible we must divide by 2 to get the correct answer.

Total Different Handshakes = n(n-1)/2

Handshake Problem as a Combinations Problem

We can also solve this Handshake Problem as a combinations problem as C(n,2).

n (objects) = number of people in the group
r (sample) = 2, the number of people involved in each different handshake

The order of the items chosen in the subset does not matter so for a group of 3 it will count 1 with 2, 1 with 3, and 2 with 3 but ignore 2 with 1, 3 with 1, and 3 with 2 because these last 3 are duplicates of the first 3 respectively.

\( C(n,r) = \dfrac{n!}{( r! (n - r)! )} \)

\( C(n,2) = \dfrac{n!}{( 2! (n - 2)! )} \)

expanding the factorials,

\( = \dfrac{1\times2\times\times(n-2)\times(n-1)\times(n)}{( 2\times1\times(1\times2\times\times(n-2)) )} \)

cancelling and simplifying,

\( = \dfrac{(n-1)\times(n)}{2} = \dfrac{n(n-1)}{2} \)

which is the same as the equation above.


[1] Zwillinger, Daniel (Editor-in-Chief). CRC Standard Mathematical Tables and Formulae, 31st Edition New York, NY: CRC Press, p. ,

For more information on combinations and binomial coefficients please see Wolfram MathWorld: Combination.


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Calculator c

Permutation and Combination Calculator


Permutations, nPr = 30
Combinations, nCr = 15

Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. A typical combination lock for example, should technically be called a permutation lock by mathematical standards, since the order of the numbers entered is important; is not the same as , whereas for a combination, any order of those three numbers would suffice. There are different types of permutations and combinations, but the calculator above only considers the case without replacement, also referred to as without repetition. This means that for the example of the combination lock above, this calculator does not compute the case where the combination lock can have repeated values, for example,


The calculator provided computes one of the most typical concepts of permutations where arrangements of a fixed number of elements r, are taken from a given set n. Essentially this can be referred to as r-permutations of n or partial permutations, denoted as nPr, nPr, P(n,r), or P(n,r) among others. In the case of permutations without replacement, all possible ways that elements in a set can be listed in a particular order are considered, but the number of choices reduces each time an element is chosen, rather than a case such as the "combination" lock, where a value can occur multiple times, such as For example, in trying to determine the number of ways that a team captain and goalkeeper of a soccer team can be picked from a team consisting of 11 members, the team captain and the goalkeeper cannot be the same person, and once chosen, must be removed from the set. The letters A through K will represent the 11 different members of the team:

A B C D E F G H I J K   11 members; A is chosen as captain

B C D E F G H I J K   10 members; B is chosen as keeper

As can be seen, the first choice was for A to be captain out of the 11 initial members, but since A cannot be the team captain as well as the goalkeeper, A was removed from the set before the second choice of the goalkeeper B could be made. The total possibilities if every single member of the team's position were specified would be 11 &#; 10 &#; 9 &#; 8 &#; 7 &#; &#; 2 &#; 1, or 11 factorial, written as 11!. However, since only the team captain and goalkeeper being chosen was important in this case, only the first two choices, 11 &#; 10 = are relevant. As such, the equation for calculating permutations removes the rest of the elements, 9 &#; 8 &#; 7 &#; &#; 2 &#; 1, or 9!. Thus, the generalized equation for a permutation can be written as:

Or in this case specifically:

11P2 =  = 11 &#; 10 =

Again, the calculator provided does not calculate permutations with replacement, but for the curious, the equation is provided below:

nPr = nr


Combinations are related to permutations in that they are essentially permutations where all the redundancies are removed (as will be described below), since order in a combination is not important. Combinations, like permutations, are denoted in various ways, including nCr, nCr, C(n,r), or C(n,r), or most commonly as simply

. As with permutations, the calculator provided only considers the case of combinations without replacement, and the case of combinations with replacement will not be discussed. Using the example of a soccer team again, find the number of ways to choose 2 strikers from a team of Unlike the case given in the permutation example, where the captain was chosen first, then the goalkeeper, the order in which the strikers are chosen does not matter, since they will both be strikers. Referring again to the soccer team as the letters Athrough K, it does not matter whether Aand then Bor Band then Aare chosen to be strikers in those respective orders, only that they are chosen. The possible number of arrangements for all npeople, is simply n!, as described in the permutations section. To determine the number of combinations, it is necessary to remove the redundancies from the total number of permutations ( from the previous example in the permutations section) by dividing the redundancies, which in this case is 2!. Again, this is because order no longer matters, so the permutation equation needs to be reduced by the number of ways the players can be chosen, Athen Bor Bthen A, 2, or 2!. This yields the generalized equation for a combination as that for a permutation divided by the number of redundancies, and is typically known as the binomial coefficient:

Or in this case specifically:

11C2 =  = 55

It makes sense that there are fewer choices for a combination than a permutation, since the redundancies are being removed. Again for the curious, the equation for combinations with replacement is provided below:

(r + n -1)!
r! &#; (n - 1)!
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Create Calculator in C programming Language in just 10 minutes with line by line code explanation

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