Saturday, March 1, 2014

Permutations

Posted by Ramandeep Singh
Permutations are very important in Quantitative Aptitude. They are most often used for finding probabilities or the number of ways something can be arranged. In fact, a permutation is simply the number of different ways a set can be arranged.

An Example of a Permutation


One of the most used examples of a permutation is the number of ways to arrange a batting order for a baseball team.

Example1:  A baseball team has 9 players, each of which go up and bat before repeating the order. How many different ways can you organize 9 players in an order?

Solution: For the first position, there would be 9 options, for the second position 8 options, for the third 7, and so on, until for the 9th position there would be only one option (Since no player can go twice before all of the players have gone). Hence, the number of positions would be

9*8*7*6*5*4*3*2*1= 362,880

From that, we know that there are 362,880 ways to order a batting order for a baseball team. Writing 9*8*7*6... can be very tedious, though. Hence, we have what we call a "factorial"

Factorials

Whenever you multiply a number by the number right before it, then by the number right before that and so on until you multiply it by one, we simply write it as "n!" where n is the number. So 9 factorial, like in the example above, would be written 9!

9!= `9 times 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1`

Permutations where the entire set isn't used

Most of the time, unlike in the baseball example, you will be selecting a few objects out of the set, not every object. The number of objects you take is called r. The size of the set is n.

It is expressed as P(n,r)

Example2: Choosing the 1st, 2nd and 3rd places out of a field of 9 players would be expressed as P(9,3), since there are 9 objects and three are being selected. How many different ways are there?

Solution: To solve this, you use the formula P(n,r) = n!/(n-r)!
 So for the top three out of nine, it would be

P(n,r) = n!/(n-r)!
P(9,3) = 9!/(9-3)!
P= 9!/6!

Now if you take that back into your old form, you get this:

P=9*8*7*6*5*4*3*2*1/6*5*4*3*2*1

Since they are all being multiplied, you can simply cancel out the same numbers on the top and bottom. Hence, you may remove the 6, 5, 4, 3, 2 and 1 from both top and bottom.

P=9*8*7
P= 504

So there are 504 ways to choose 1st 2nd and 3rd place out of 9 players.

Wait, that seems harder than it should be...

It is. Really, you can skip the whole P=n!/(n-r)! and just multiply the first r objects. so instead of that, you just do P=9*8*7. It's much simpler. Even so, you should remember the formula for tests.

What to remember:

  • A permutation is the number of ways objects can be ordered in a set. Order DOES matter.

  • When the whole set is used, it is simply n!

  • n! means n factorial, or n*(n-1)*(n-2)... down to 1.

  • When you only choose some of the set, simply take the number you are supposed to choose and use that many numbers from the start of the factorial. So P(9,3)=9*8*7 and P(121,4)=121*120*119*118

  • Remember the formula P=n!/(n-r)! for tests.

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