Would I use combinations or permutations?If 6 people are running in a race, how many possible ways can they come in if there are no ties and everyone finishes the race?

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mcetner's profile pic

mcetner | High School Teacher | (Level 1) Adjunct Educator

Posted on

The easiest way to remember the difference between combinations and permutations is that for permutations, order matters, and for combinations, order does not matter.  Since, when you are runing in a race, it matters what order you come in, you would be using permutations. 

You have 6 choices for first place, then 5 for second place, 4 for third place, etc.  So, the number of ways they can come in is 6*5*4*3*2*1m or 6!.  This is the same as 6p6. 

Note:  If you only wanted how many ways they could come in first place, as the question originally asked, it would be 6p1, because you only want 1 place.

atyourservice's profile pic

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

combinations because we are looking at the number of options and since we are

multiplying no matter what order the numbers are in the answer will still be the

same. Just do

6 x 5 x 4 x 3 x 2 x 1 = 720

revolution's profile pic

revolution | College Teacher | (Level 1) Valedictorian

Posted on

Combinations have no regard to order but permutations however, has regard to the order of pattern

Lets look at the equation. Let the six runners be named as A,B,C,D,E and F respectively.

A has six choices, while B has 6-1 choices if A came first. C might get 6-2 chances if A and B get first and second place. D, E and F follows in the same way as A and B.

So, 6P6 or 6!= 6*5*4*3*2*1= 720 chances

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Permutation is as good as arranging. Whereas,in selecting a group of persons, we are interested only in number of persons irrespective of the order of selection. So we use combinations .

Let A,B,C,D,E,F be the 6 persons in race.

A may come in any of the 6 places. So, 6 choice.

For any one place of A, B can occupy any one of the other 5 places.

For any particular order of A and B, C ha the choice of the places  6-2=4 places.

Similarly D has the choice of 6-3= 3places.

E can occupy 2 places and F has to be at the available one place.

Thus, the choces being free and indepedent, the number of aarrangements or ways is 6P6 given by:

6P6 = 6(6-1)(6-2)(6-3)(6-4)(6-5) = 6! = 720

 

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