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Unique combinations between 246 countries and 7,000 cities

by Janice Tetlock
(Vancouver BC)











































How many combos if there are 246 countries and 7,000 cities?

We developed and App that places calls between two countries.

There are 246 countries and 7,000 possible cities to call.

How many combos if there are 246 countries and 7,000 cities?

How many unique combinations does that make?

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Nov 11, 2012
Unique Combinations
by: Staff


Answer

Part I


Just to make sure there is no confusion over terminology, here are two definitions which are important:


PERMUTATIONS apply to lists where the order of the possible choices does matter.


For example, suppose you have a list of three numbers: 1, 2, and 3. You know that these three numbers, arranged in the proper order, can be used to unlock a safe. You also know that you must use all three numbers and you cannot use each number more than once.

There are six possible permutations because the same three numbers can be arranged in six different of ways:


{1,2,3}
{1,3,2}
{2,1,3}
{2,3,1}
{3,1,2}
{3,2,1}


However, you do not need to list every possible choice to know there are six possible ways to arrange the numbers.

The number of possible ways to arrange the numbers 1, 2, and 3 can be calculated as follows:

Choice of 1st number: 3 possible choices (1, 2, or 3)

Choice of 2nd number: 2 possible choices (the two numbers remaining after the 1st number was chosen)

Choice of 3rd number: 1 possible choice (the last number remaining)

Permutations = 3*2*1

Permutations = 6


The general formula is:

           Permutations = n!/[(n - r)!]

n = 3 (the number of possible choices)

r = 3 (the number of choices made at one time)

Permutations = 3!/[(3 - 3)!]

Permutations = 3!/(0!)

Permutations = (3*2)/1

Permutations = 6


COMBINATIONS apply to groups. How many different groups are there? The Order of the elements within a group does NOT matter.


Suppose you have a GROUP of three numbers: 1, 2, and 3. (Remember, this is a group of three numbers, NOT A LIST of three numbers)

There is only one combination of the three numbers since order of the numbers does not matter.

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Nov 11, 2012
Unique Combinations
by: Staff

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Part II

{1,2,3} is just as valid as {2,3,1}. Both groups use the same three numbers.


Now suppose you have a group four numbers: 1, 2, 3, and 4

Also suppose that you could only choose 3 numbers at a time.

There are 4 possible combinations because those four numbers can be combined in 4 different groups:

{1,2,3}
{1,2,4}
{1,3,4}
{2,3,4}

However, you do not need to list every possible choice to know there are 4 possible ways to combine the numbers.

The number of possible ways to arrange the numbers 1, 2, 3, and 4 can be calculated as follows:


Choice of 1st number: 4 possible choices (1, 2, 3, or 4)

Choice of 2nd number: 3 possible choices (the three numbers remaining after the 1st number was chosen)

Choice of 3rd number: 2 possible choices (the last two numbers remaining)

Since only three numbers will be chosen, the last number (whatever it is) will not be included.

The general formula is:

           Combinations = Permutations ÷ (r!)
           (how many different groups)

           Combinations = n!/[(n - r)!(r!)]
           (how many different groups)

n = 4 (the number of possible choices)

r = 3 (the number of choices made at one time)

Combinations = 4!/[(4 - 3)!(3!)]

Combinations = 4!/[(1!)(3!)]

Combinations = (4*3*2*1)/[(1)(3*2*1)]

Combinations = 4


That said, let’s look at your problem statement:

We developed and App that places calls between two countries.

There are 246 countries and 7,000 possible cities to call.

How many combos if there are 246 countries and 7,000 cities?

How many unique combinations does that make?

-------------------------------------------

Nov 11, 2012
Unique Combinations
by: Staff

-------------------------------------------

Part III


Your problem statement does not specify if the App only places calls between two different countries, or between two different cities, or between any two cities (including calls between two people in the same city).

If the App only routes calls between two different countries at a time, the number of possible combinations is:

Combinations = n!/[(n - r)!(r!)]

n = 246 (the number of possible choices)

r = 2 (the number of choices made at one time)

Combinations = 246!/[(246 - 2)!(2!)]

Combinations = 246!/[(244)!(2!)]

Combinations = (246 * 245 * 244!)/[(244)!(2!)]

Combinations = (246 * 245)/(2 * 1)

Combinations = 30,135


If the App can route calls between any two different cities at one time , the number of possible combinations is:


Combinations = n!/[(n - r)!(r!)]

n = 7000 (the number of possible choices)

r = 2 (the number of choices made at one time)

Combinations = 7000!/[(7000 - 2)!(2!)]

Combinations = (7000 * 6999 * 6998!)/[(6998)!(2!)]

Combinations = (7000 * 6999)/(2!)

Combinations = (7000 * 6999)/(2 * 1)

Combinations = 24,496,500


If the App can route calls between phones in any of the cities at one time, (the caller can place a call to someone in the same city as the caller, or to a different city), the number of possible combinations is:


Combinations = (n + r – 1)!/[(n - r)!(r!)]

n = 7000 (the number of possible choices)

r = 2 (the number of choices made at one time)

Combinations = (7000 + 2 – 1)!/[(7000 - 2)!(2!)]

Combinations = (7001!)/[(6998)!(2!)]

Combinations = (7001 * 7000 * 6999 * 6998!)/[(6998)!(2!)]

Combinations = (7001 * 7000 * 6999)/(2!)

Combinations = (7001 * 7000 * 6999)/(2 * 1)

Combinations = 171,499,996,500




Thanks for writing.

Staff
www.solving-math-problems.com


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