In law entrance tests, questions from Permutation and Combination cover a good part (around 3-4 questions) of math’s section.
These questions are mostly easy but only if one has her basic concepts clear. So in this article an attempt has been made to do so.
So starting with permutation, a permutation is an arrangement of all or part of a number of things in a definite order.
For example, the permutations of the three letters a, b, c taken all at a time are:
abc, acb, bca, bac, cba, cab. The permutations of the three letters a, b, c taken two at a time are- ab, ac, ba, bc, ca, cb.
Thus 8P 3 denotes the numbers of permutations of 8 different things taken 3 at a time and 5P 5 denotes the number of permutations of 5 different things taken 5 at a time. Important In permutations, the order of arrangement is taken into account; when the order is changed, a different permutation is obtained.
How many ways can we award a 1st, 2nd and 3rd place prize among eight contestants? (Gold / Silver / Bronze)
We’re going to use permutations since the order we hand out these medals matters.
Here’s how it breaks down:
Gold medal: 8 choices: A B C D E F G H. Let’s say A wins the Gold.
Silver medal: 7 choices: B C D E F G H. Let’s say B wins the silver.
Bronze medal: 6 choices: C D E F G H. Let’s say… C wins the bronze.
We picked certain people to win, but the details don’t matter: we had 8 choices at first, then 7, then 6.
We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.
Where 8!/5! is just a fancy way of saying “Use the first 3 numbers of 8!”. If we have n items total and want to pick k in a certain order, we get: P(n, k) or nPk = n! / (n – k)! (general formula)
So ways in which we can award 1 st 2 nd 3 rd price will be = 8.7.6 = 336
If we find the total number of ways in which 4 persons can take their places in a cab having 6 seats.
The number of ways in which 4 persons can take their places in a cab having 6 seats
6p4 = 6 × 5 × 4 × 3 = 360 ways.
Mario, Sandy, Fred, and Shanna are running for the offices of president, secretary and treasurer. So, in how many ways can these offices be filled: 4P3 = 4 x 3 x 2 = 24. The offices can be filled 24 ways.
Coming on Combinations, these are easy going. Order doesn’t matter.
The general formula is:
C(n, k) or nCk = P(n, k) / k!
Which means “Find all the ways to pick k people from n, and divide by the k! variants”. Writing this out, we get our combination formula, or the number of ways to combine k items from a set of n:
C(n, k) = n! / (n-k)!.k!
A few examples
Here’s a few examples of combinations (order doesn’t matter) from permutations (order matters).
Combination: Picking up 3 pens from a bunch of 10.
C(10,3) = 10!/(7!·3!) = 10 · 9 · 8 / (3 · 2 · 1) = 120.
Permutation: Picking a President, VP and Secretary from a group of 10.
P (10,3) = 10!/7! = 10 · 9 · 8 = 720.
Combination: Choosing 3 dishes from a menu of 10.
C(10,3) = 120.
Permutation: Listing your 3 favorite dishes, in order, from a menu of 10.
P(10,3) = 720.
The difficulty arises sometimes when we fail to determine if the situation represents a permutation or a combination. So we need to firstly determine that order is important or not in given situation.
In a Permutation the order of occurrence of the object or the arrangement is important but in Combination the order of occurrence of the object or the arrangement is not important.
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