In mathematics, permutation refers to the arrangement of all the members of a set in some order or sequence, while combination does not regard the order as a parameter. It is just a way of selecting items from a set or collection.
Permutation Formula:Â A permutation is the arrangements ofÂ rÂ things from a set ofÂ nÂ things without replacement. Order matters in the permutation.
\(nP_{r}=\frac{n!}{(n-r)!}\)
Combination Formula:Â A combination is the choice ofÂ rÂ things from a set ofÂ nÂ things without replacement. The order does not matter in combination.
\(nC_{r}=\frac{n!}{(n-r)!r!}=\frac{nPr}{r!}\)
Derivation:
- Number of permutations of n different things taking r at a time isÂ nPr.
Let us assume that there are r boxes and each of them can hold one thing. There will be as many permutations as there are ways of filling inÂ rÂ vacant boxes byÂ nÂ objects.
No. of ways the first box can be filled:Â n
No. of ways the second box can be filled: (nÂ â€“ 1)
No. of ways the third box can be filled: (nÂ â€“ 2)
No. of ways the fourth box can be filled: (nÂ â€“ 3)
No. of waysÂ rthÂ box can be filled: (nÂ â€“ (rÂ â€“ 1))
Therefore, no. of ways of filling inÂ rÂ boxes in succession can be given by:
nÂ (nÂ â€“ 1) (nÂ â€“ 2) (n-3) . . . (nÂ â€“ (rÂ â€“ 1))
This can be written as:
nÂ (nÂ â€“ 1) (nÂ â€“ 2) â€¦ (nÂ â€“Â rÂ + 1)
The no. of permutations ofÂ nÂ different objects takenÂ rÂ at a time, where 0 <Â rÂ â‰¤Â nÂ and the objects do not repeat is:
nÂ (nÂ â€“ 1) (nÂ â€“ 2) (nÂ â€“ 3) . . . (nÂ â€“Â rÂ + 1).
â‡’ Â Â nPrÂ = n ( n â€“ 1) ( n â€“ 2)(Â n â€“ 3). . .( n â€“ r + 1)
Multiplying and divided by (nÂ â€“Â r) (nÂ â€“Â rÂ â€“ 1) . . . 3 Ã— 2 Ã— 1, we get
\(nP_{r}=\frac{[n(n-1)(n-2)(n-3)â€¦(n-r+1)(n-r)(n-r-1)..3\times 2\times 1]}{(n-r)(n-r-1)..3\times 2\times 1}=\frac{n!}{(n-r)!}\)
Hence,
\(nP_{r}=\frac{n!}{(n-r)!}\)
WhereÂ 0 < r â‰¤ n
- Number of combinations of n distinct things taking r at a time is Â nCr.
No. of ways to select the first object fromÂ nÂ distinct objects:Â nÂ ways
No. of ways to select the second object from (n-1)Â distinct objects: (n-1)Â ways
No. of ways to select the third object from (n-2)Â distinct objects: (n-2)Â ways
No. of ways to select the fourth object from (n-3)Â distinct objects: (n-3)Â ways
No. of ways to select rthÂ object from (n-(r-1))Â distinct objects: (n-(r-1))Â ways
Completing selectionÂ rÂ things of the original set of nÂ things creates an orderedÂ sub-setÂ ofÂ rÂ elements.
âˆ´ The number of ways to make a selection of rÂ elements of the original set ofÂ nÂ elements isÂ nÂ (nÂ â€“ 1) (nÂ â€“ 2) (n-3) . . . (nÂ â€“ (rÂ â€“ 1)) orÂ nÂ (nÂ â€“ 1) (nÂ â€“ 2) â€¦ (nÂ â€“Â rÂ + 1)
Let us consider the ordered sub-setÂ ofÂ rÂ elements andÂ allÂ itsÂ permutations. The total number of all permutations of this sub-set is equal to r!Â becauseÂ rÂ objects in every combination can be rearranged inÂ r!Â ways.
Hence, the total number of permutations ofÂ nÂ different things takenÂ rÂ at a time isÂ nCrÂ Ã—r!Â On the other hand, it isÂ nPr.
\(nP_{r}=nC_{r}\times r!\)
\(nC_{r}=\frac{nP_{r}}{r!}=\frac{n!}{(n-r)!r!}\)
Permutations and Combinations in Real Life
Permutations and combinationsÂ areÂ techniques which help us to answer the questions or determine the number of different ways of arranging and selecting objects without actually listing them in real life. For example, when you have to arrange people, pick a team captain, pick two favorite colors, in order, from a color brochure, or selection of menu, food, clothes, subjects, team, etc.
Solved Examples
Questions 1:Â Evaluate
- 12P2
- 10C3
Solution:
- Here, n = 12 and r = 2
\(12P_{2}=\frac{12!}{(12-2)!} = 132\)
- Here, n = 10 and r = 3
\(10C_{3}=\frac{10!}{(10-3)!3!}=\frac{10!}{7!3!}=120\)
Questions 2:Â Teacher asks a student to choose 6 items from the table. If the table has 20 items to choose, how many ways could the students choose the things?
Solution:Â Here, the student has to choose 6 items from 20 items.
Here, r= 6 and n= 20
Combination,
\(nC_{r}=\frac{n!}{(n-r)!r!}\)
\(20C_{6}=\frac{20!}{(20-6)!6!} = 38760\)