How can I choose which to use permutation or combination to solve a given statistical problem in maths. Eg. In how many ways can the letters of word: "ARRANGE" be arranged? How many of these arrangements are there in which a) the two R's together b)the two R's and two A's come together
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If you do a search on this website for "ARRANGE" you'll find lots of questions like this – Martin Hansen Mar 01 '19 at 18:12
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@MartinHansen thankyou martin the answer is correct ,but my basic question is if there is a way to decide methods from permutation and combination when solving such problems just by looking – Nahom Mar 01 '19 at 18:25
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I've added an explanation - hope it makes sense. – Martin Hansen Mar 01 '19 at 21:02
2 Answers
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The answer to the first question is $\frac {7!} {2! \cdot 2!} = 1260$ and the answer to $(a)$ is $\frac {6!} {2!} = 360$ and answer to $(b)$ is $5! = 120.$
little o
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2Although correct, it helps to explain why each number was chosen and the thought process on how to approach the problem. – JMoravitz Mar 01 '19 at 18:23
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Part b) is easier so I'll do that first.
Imagine you have 5 cards on which is written $$ aa : rr : n : g : e$$ You are going to place these cards is a line on a table, There are 5 possible cards you could place first, then four second, three third, two fourth and then what ever card is left in your hand goes down last. So that's, $$5 \times 4 \times 3 \times 2 \ 1 = 5! = 120$$ With part a) you have 6 cards on which is written $$ a : a : rr : n : g : e$$ So repeating the argument above there are 6! ways of placing the cards. There is a catch however which is the two 'a' cards that look the same. So you can't tell the difference between, for example, $$n \implies a_1 \implies rr \implies g \implies a_2 \implies e$$ and $$n \implies a_2 \implies rr \implies g \implies a_1 \implies e$$ so the 6! needs dividing by 2!, i.e. 2, giving an answer of 360
Hope that helps !
Martin Hansen
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