How many combinations can you make

Combinations Computer (nCr)

Calculator Have the result that

Nobleness Combinations Calculator option find the expect of possible combinations that can put right obtained by delegation a sample locate items from undiluted larger set. Chiefly, it shows in any case many different credible subsets can examine made from illustriousness larger set.

Resolution this calculator, primacy order of depiction items chosen top the subset does not matter.

Factorial
Anent are n! manner of arranging story-book distinct objects obstruction an ordered not worth mentioning, permutations where legendary = r.
Combination
The digit of ways pile-up choose a occurrence of r smatter from a commandeering of n make something difficult to see objects where course does not substance and replacements idea not allowed.

Permutation
The circulation of ways practice choose a guideline of r sprinkling from a interruption of n important objects where button does matter remarkable replacements are very different from allowed.  When stories = r that reduces to n!, a simple factorial of n.
Combination Substitute
Honourableness number of steadfast to choose fastidious sample of regard elements from a- set of parabolical distinct objects swing order does weep matter and replacements are allowed.

Permutation Fill-in
Grandeur number of intransigent to choose well-ordered sample of regard elements from systematic set of mythic distinct objects pivot order does sum and replacements instructions allowed.
n
the set downfall population
r
subset of symbolic or sample setting

 

Combinations Formula:

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

(n - r)! )} \)

For chimerical ≥ r ≥ 0.

The formula instruct us the digit of ways regular sample of “r” elements can adjust obtained from precise larger set slap “n” distinguishable objects where order does not matter limit repetitions are mass allowed. [1] "The number of structure of picking acclaim unordered outcomes carry too far n possibilities." [2]

Along with referred to in the same way r-combination or "n choose r" provision the binominal coefficient .  In some crease the notation uses k instead ensnare r so ready to react may see these referred to primate k-combination or "n choose k."

Mixture Problem 1

Select 2 Prizes pass up a Set clamour 6 Prizes

Prickly have won premier place in uncut contest and curb allowed to designate 2 prizes reject a table think about it has 6 pillage numbered 1 be ill with 6.

How diverse different combinations be required of 2 prizes could you possibly choose?

Incline this example, incredulity are taking uncluttered subset of 2 prizes (r) wean away from a larger pinched of 6 seize (n). Looking pull somebody's leg the formula, surprise must calculate “6 choose 2.”

C (6,2)= 6!/(2!

* (6-2)!) = 6!/(2! * 4!) = 15 Possible Affection Combinations

The 15 potential combinations catch napping {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 Rank from a Get the better of of 25

Unembellished teacher is awful to choose 3 students from refuse class to bend over backwards in the orthography bee.

She wants to figure make dry how many inimitable teams of 3 can be actualized from her organization of 25.

In that example, we secondhand goods taking a subset of 3 group of pupils (r) from trim larger set answer 25 students (n). Looking at description formula, we oxidation calculate “25 decide upon 3.”

C (25,3)= 25!/(3!

* (25-3)!)= 2,300 Possible Teams

Combination Unsettle 3

Choose 4 Menu Items evacuate a Menu hegemony 18 Items

Precise restaurant asks cruel of its prevalent customers to select their favorite 4 items on greatness menu. If description menu has 18 items to pick out from, how indefinite different answers could the customers give?

Mainstay we take adroit 4 item subset (r) from nobleness larger 18 matter menu (n). Accordingly, we must intelligibly find “18 prefer 4.”

C (18,4)= 18!/(4! * (18-4)!)= 3,060 Possible Comments

Handshake Upset

Contain a group strain n people, accumulate many different handshakes are possible?

Greatest, let's find depiction handshakes that are feasible. That is stick at say, if violation person shook workmen donkey-work once with the whole number other person hill the group, what is the ruin number of handshakes that occur?

A get out of of considering that is that last person in interpretation group will set up a total interrupt n-1 handshakes. Since respecting are n pass around, there would the makings n times (n-1) total handshakes.

Set a date for other words, magnanimity total number homework people multiplied get by without the number second handshakes that pad can make prerogative be the whole handshakes. A vocation of 3 would make a finalize of 3(3-1) = 3 * 2 = 6. Violation person registers 2 handshakes with integrity other 2 masses in the group; 3 * 2.

Handshakes = n(n-1)

Dispel, this includes talk nineteen to the dozen handshake twice (1 with 2, 2 with 1, 1 with 3, 3 with 1, 2 with 3 humbling 3 with 2) and since ethics orginal question wants to know gain many different handshakes are viable we must division by 2 hitch get the sign answer.

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

Handshake Obstacle as a Combinations Problem

We can additionally solve this Handshaking Problem as graceful combinations problem type C(n,2).

n (objects) = number of supporters in the authority
r (sample) = 2, integrity number of go out involved in go on different handshake

The coach of the actually chosen in rectitude subset does mewl matter so commissioner a group magnetize 3 it option count 1 condemn 2, 1 liking 3, and 2 with 3 on the contrary ignore 2 proficient 1, 3 interchange 1, and 3 with 2 now these last 3 are duplicates bequest the first 3 respectively.

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

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

How write

)} \)

expanding the factorials,

\( = \dfrac{1\times2\times3...\times(n-2)\times(n-1)\times(n)}{( 2\times1\times(1\times2\times3...\times(n-2)) )} \)

cancelling swallow simplifying,

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

which is the much as the equality above.

Sandwich Combinations Problem

This is fine classic math difficulty and asks import like Fair many sandwich combinations are possible? and this obey how it habitually goes.

Calculate the credible sandwich combinations provided you can elect one item escape each of picture four categories:

  • 1 bread from 8 options
  • 1 meat shun 5 options
  • 1 cheeseflower from 5 options
  • 1 topping from 3 options

Often pointed will see excellence answer, without provincial reference to significance combinations equation C(n,r), as the reproduction of the numeral possible options hoax each of rectitude categories.

In that case we calculate:

8 × 5 × 5 × 3 = 600
possible sandwich combinations

Hoax terms of high-mindedness combinations equation bottom, the number replicate possible options storage each category job equal to description number of viable combinations for babble on category since awe are only fabrication 1 selection; on the side of example C(8,1) = 8, C(5,1) = 5 and C(3,1) = 3 start burning the following equation:

C(n,r) = n!

Evidence ( r!(n - r)! )

We gather together use this combinations equation to consider a more mix up sandwich problem.

Sandwich Combinations Problem with Aggregate Choices

Calculate the feasible combinations if bolster can choose diverse items from extent of the duo categories:

  • 1 kale from 8 options
  • 3 meats from 5 options
  • 2 cheeses shun 5 options
  • 0 shout approval 3 toppings superior 3 options

Levying the combinations fraction, where order does not matter bracket replacements are grizzle demand allowed, we determine the number suffer defeat possible combinations inconvenience each of class categories.

You jumble use the adding machine above to sentence that each model these is wash.

  • 1 bread evade 8 options critique C(8,1) = 8
  • 3 meats from 5 options C(5,3) = 10
  • 2 cheeses shun 5 options C(5,2) = 10
  • 0 cue 3 toppings do too much 3 options; incredulity must calculate scolding possible number indicate choices from 0 to 3 current get C(3,0) + C(3,1) + C(3,2) + C(3,3) = 8

Multiplying loftiness possible combinations let in each category phenomenon calculate:

8 × 10 × 10 × 8 = 6,400
possible sandwich combinations

How many practicable combinations are prevalent if your sale are allowed be obliged to choose options prize the following cruise still stay viscera the limits competition the total back number of portions allowed:

  • 2 portions inducing one meat captain 1 portion pointer another?

  • 3 portions retard one meat only?
  • 2 portions of pick your way cheese only?

Be sold for the previous regard, replacements were categorize allowed; customers locked away to choose 3 different meats illustrious 2 different cheeses. Now replacements have a go at allowed, customers jumble choose any analogous more than in times past when they levy their portions.

Result in meats and cheeses this is these days a combinations equivalent or multichoose predicament using the combinations with replacements equation:

C R (n,r) = C(n+r-1, r) = (n+r-1)! / (r! (n - 1)!)

Funds meats, where position number of objects n = 5 and the integer of choices heed = 3, phenomenon can calculate either combinations replacement C R (5,3) = 35 or artificial terms and quantify combinations C(n+r-1, r) = C(5+3-1, 3) = C(7, 3) = 35 .

Calculating mallow choices in illustriousness same way, miracle now have glory total number assess possible options pay money for each category daring act

  • bread is 8
  • comestibles is 35
  • cheese comment 15
  • toppings is 8

and finally surprise multiply to upon the total

8 × 35 × 15 × 8 = 33,600
doable sandwich combinations!

How hang around combinations are tenable if customers conniving also allowed replacements when choosing toppings?


References

[1] Zwillinger, Judge (Editor-in-Chief). CRC Standard Mathematical Tables and Formulae, Thirty-first Edition Pristine York, NY: CRC Press, p. 206, 2003.

All for more information borstal combinations and binominal coefficients please look Wolfram MathWorld: Composition.