![]() ![]() Now, for each of these scenarios, now, so we have 20 scenarios. So, for each of those scenarios, four different peopleĬould sit in chair two. If we seated people in order, which we might as well do, we could say, look, five different people could sit in chair one. People in these five chairs, we could say, well, we The number of scenarios, or we could say the number of permutations of putting these five Two, position three, position four, and position five. Them in five different, let's say, positions, or chairs, so position one, position If we had five people, let's say person A, person B, person C, person D, and person E, and we wanted to put ) so we need to multiply 24 by 6 to get our answer.Ģ4*6=144 ways to arrange pencil so that "p e and n" are always next to each other !Ĭan you come up with a formula for a "n" long word where you want "x" items to be together ? :) Let's take a second and figure out what 24 actually represents, 24 is the number of ways you can arrange "pencil" with "pen" being a single letter, but we got 6 of those (pen, nep, epn. And "pencil" just became a 4 letters word, so it has 4! ways of being arranged, therefore 24 ways. Now we're doing it with 3 letters "p e and n" so there are 3! ways of arranging them, therefore 6 ways. Finally we added them together to get our answer. But really we looked at all the different ways to arrange 2 items, and there was 2! ways of doing it, so 2 ways and "pencil" was treated as a 5 letters word. With "e" and "n" we figured out that "en" and "ne" were the only two possible way to arrange them, and treated them as a single letter. Now to dig in a little deeper what if we wanted "p e and n" to always be together ? Have a go at it before looking below :) There we go ! There are 240 different ways to arrange "pencil" so that e and n are always next to each other. Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance.First let's think about it a bit, to figure out how many ways you can arrange "Pencil" with N and E always next to one another it's going to be all the different ways to arrange "pencil" as if "en" was a single letter and the same thing with "ne"ġ) How many ways can we arrange Pencil as if "en" was a single letter?ĥ! = 120 ways, we have 5 things to arrange P c i l and "en"Ģ) Now how many ways can we arrange Pencil as if "ne" was a single letter? The number of ways of choosing 6 numbers from 49 is 49C 6 = 13 983 816. What is the probability of winning the National Lottery? You win if the 6 balls you pick match the six balls selected by the machine. ![]() In the National Lottery, 6 numbers are chosen from 49. The above facts can be used to help solve problems in probability. There are therefore 720 different ways of picking the top three goals. ![]() Since the order is important, it is the permutation formula which we use. In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. The number of ordered arrangements of r objects taken from n unlike objects is: How many different ways are there of selecting the three balls? There are 10 balls in a bag numbered from 1 to 10. The number of ways of selecting r objects from n unlike objects is: Therefore, the total number of ways is ½ (10-1)! = 181 440 How many different ways can they be seated?Īnti-clockwise and clockwise arrangements are the same. When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)! The number of ways of arranging n unlike objects in a ring when clockwise and anticlockwise arrangements are different is (n – 1)! There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are: In how many ways can the letters in the word: STATISTICS be arranged? The number of ways of arranging n objects, of which p of one type are alike, q of a second type are alike, r of a third type are alike, etc is: The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4! The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. ![]() The second space can be filled by any of the remaining 3 letters. The first space can be filled by any one of the four letters. This is because there are four spaces to be filled: _, _, _, _ How many different ways can the letters P, Q, R, S be arranged? The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). This section covers permutations and combinations. ![]()
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