How many different ways can 4 children be ordered in a line?

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This section covers permutations and combinations.

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Arvarying Objects

The number of means of arvarying n unchoose objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1

Example

How many different ways deserve to the letters P, Q, R, S be arranged?

The answer is 4! = 24.

This is bereason there are four spaces to be filled: _, _, _, _

The initially room have the right to be filled by any kind of one of the 4 letters. The second room have the right to be filled by any kind of of the remaining 3 letters. The 3rd area can be filled by any of the 2 staying letters and also the last room have to be filled by the one remaining letter. The total variety of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!

The variety of methods of arranging n objects, of which p of one kind are alike, q of a second kind are aprefer, r of a third form are aprefer, and so on is:

n! .p! q! r! …

Example

In just how many ways have the right to the letters in the word: STATISTICS be arranged?

There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arvarying the letters are:

10!=50 4003! 2! 3!

Rings and Roundabouts

The variety of means of arranging n unchoose objects in a ring once clockwise and also anticlockwise arrangements are different is (n – 1)!

When clockwise and anti-clockwise arrangements are the exact same, the number of methods is ½ (n – 1)!

Example

Ten people go to a party. How many kind of various means deserve to they be seated?

Anti-clockwise and also clockwise arrangements are the exact same. Thus, the full number of ways is ½ (10-1)! = 181 440

Combinations

The variety of methods of selecting r objects from n unprefer objects is:

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Example

Tright here are 10 balls in a bag numbered from 1 to 10. Three balls are schosen at random. How many type of various means are tbelow of choosing the three balls?

10C3 =10!=10 × 9 × 8= 120 3! (10 – 3)!3 × 2 × 1

Permutations

A permutation is an ordered arrangement.

The number of ordered arrangements of r objects taken from n unfavor objects is:

nPr = n! . (n – r)!

Example

In the Match of the Day’s goal of the month competition, you had actually to pick the optimal 3 purposes out of 10. Because the order is essential, it is the permutation formula which we usage.

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10P3 =10! 7!

= 720

Tbelow are therefore 720 different methods of picking the peak 3 goals.

Probability

The above facts have the right to be used to help deal with troubles in probcapacity.

Example

In the National Lottery, 6 numbers are liked from 49. You win if the 6 balls you pick match the six balls selected by the machine. What is the probcapability of winning the National Lottery?

The variety of ways of picking 6 numbers from 49 is 49C6 = 13 983 816 .

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Because of this the probcapability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is around a 1 in 14 million possibility.