Find the number of permutations and combinations if n = 12 and r = 2
Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter.
In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered.
Permutation Formula
In permutation r things are picked from a group of n things without any replacement. In this order of picking matter.
nPr = (n!)/(n – r)!
Here,
n = group size, the total number of things in the group
r = subset size, the number of things to be selected from the group
Combination
A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, the order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used.
Combination Formula
In combination r things are picked from a set of n things and where the order of picking does not matter.
nCr = n!⁄((n-r)! r!)
Here,
n = Number of items in set
r = Number of things picked from the group
Find the number of permutations and combinations if n = 12 and r = 2.
Solution:
Permutation Formula = nPr
nPr = n!/(n-r)!
= 12P2
= 12!/(12-2)!
= 12!/10!
= 12 × 11 × 10!/10!
= 132
Combination Formula = nCr
nCr = n!/(n-r)!r!
= 12C2
= 12!/(12-2)!2!
= 12!/10!2!
= 12×11×10!/10!2!
= 12×11/2
= 6 × 11
= 66
Similar Questions
Question 1: Find the number of permutations and combinations if n = 10 and r = 2
Solution:
Permutation Formula = nPr
nPr = n!/(n-r)!
= 10P2
= 10!/(10-2)!
= 10!/8!
= 10×9×8!/8!
= 90
Combination Formula = nCr
nCr = n!/(n-r)!r!
= 10C2
= 10!/(10-2)!2!
= 10!/8!2!
= 10×9×8!/8!2!
= 10×9/2
= 5×9
= 45
Question 2: Find the number of permutations and combinations if n = 20 and r = 4
Solution:
Permutation Formula = nPr
nPr = n!/(n-r)!
= 20P4
= 20!/(20-4)!
= 20!/16!
= 20×19×18×17×16!/16!
= 116280
Combination Formula = nCr
nCr = n!/(n-r)!r!
= 20C4
= 20!/(20-4)!4!
= 20!/16!4!
= 20×19×18×17×16!/16!4!
= 20×19×18×17/4×3×2
= 5×19×3×17
= 4845
Question 3: Find the number of permutations and combinations if n = 22 and r = 8
Solution:
Permutation Formula = nPr
nPr = n!/(n-r)!
= 22P8
= 22!/(22-8)!
= 22!/14!
= 22×21×20×19×18×17×16×15×14!/14!
= 12893126400
Combination Formula = nCr
nCr = n!/(n-r)!r!
= 22C8
= 22!/(22-8)!8!
= 22!/14!8!
= 22×21×20×19×18×17×16×15×14!/14!×8×7×6×5×4×3×2
= 22×21×20×19×18×17×16×15/8×7×6×5×4×3×2
= 11×5×19×3×17×2×3
= 319770
Question 4: Find the number of permutations and combinations if n = 8 and r = 2
Solution:
Permutation Formula = nPr
nPr = n!/(n-r)!
= 8P2
= 8!/(8-2)!
= 8!/6!
= 8×7×6!/6!
= 56
Combination Formula = nCr
nCr = n!/(n-r)!r!
= 8C2
= 8!/(8-2)!2!
= 8!/6!2!
= 8×7×6!/6!2!
= 8×7/2
= 4×7
= 28
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