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amt 65 (1) 2009 3 Paul Scott Adelaide, SA <[email protected]> The daffy secretary A secretary has n letters and n addressed envelopes. Instead of matching each letter with the corresponding envelope, she inserts the letters in a random manner. What are the chances that every letter will be in the wrong envelope? To solve a problem like this, it is often a good idea to look first at a simple case. So, let us suppose that n = 3. Then letters A, B, C can be placed in envelopes 1, 2, 3 in the following ways. We see that there are 6 = 3! possible ways of placing the letters. If we assume that the correct matching is 1 – A, 2 – B, 3 – C, then there are just two ways for which every letter is in the wrong envelope (which two?). Thus the answer to our question is two chances in six, or 2 6 . 1. Repeat the above working for n = 4. You might also try to obtain a result for n = 5, but preferably not by constructing an immense table! What shortcuts can you devise? Let W(n) denote the probability (chance) of putting n letters in n wrong envelopes. You should have obtained for n = 4, W(4) = 9 24 ways, and for n = 5, W(5) = 44 120 ways. We might also observe that W(2) = 1 2 and W(1) = 0. There is in fact a pretty pattern here, although you could be excused for not noticing it! We note first that 6 = 3.2.1 = 3!, 24 = 4!, and 120 = 5! Now let’s see what happens when we take the difference of each successive pair of results: Then our calculations suggest that: Envelope 1 A A B B C C Envelope 2 B C A C A B Envelope 3 C B C A B A -MAIL ORDER

Australian Mathematics Teacher vol. 65 no. 1 - ERIC · Pascal’s triangle. Each row is obtained from the row above by a simple rule. Can you see what it is? What would the next row

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Page 1: Australian Mathematics Teacher vol. 65 no. 1 - ERIC · Pascal’s triangle. Each row is obtained from the row above by a simple rule. Can you see what it is? What would the next row

amt 65 (1) 2009 3

Paul ScottAdelaide, SA<[email protected]>

The daffy secretary

A secretary has n letters and n addressedenvelopes. Instead of matching each letter withthe corresponding envelope, she inserts theletters in a random manner. What are thechances that every letter will be in the wrongenvelope?

To solve a problem like this, it is often a goodidea to look first at a simple case. So, let ussuppose that n = 3. Then letters A, B, C can beplaced in envelopes 1, 2, 3 in the following ways.

We see that there are 6 = 3! possible ways ofplacing the letters. If we assume that the correctmatching is 1 – A, 2 – B, 3 – C, then there arejust two ways for which every letter is in thewrong envelope (which two?). Thus the answerto our question is two chances in six, or 2

6.

1. Repeat the above working for n = 4. Youmight also try to obtain a result for n = 5,but preferably not by constructing animmense table! What shortcuts can youdevise?

Let W(n) denote the probability (chance) ofputting n letters in n wrong envelopes. Youshould have obtained for n = 4, W(4) = 9

24 ways,and for n = 5, W(5) = 44

120 ways. We might alsoobserve that W(2) = 1

2 and W(1) = 0. There is infact a pretty pattern here, although you couldbe excused for not noticing it! We note first that6 = 3.2.1 = 3!, 24 = 4!, and 120 = 5! Now let’ssee what happens when we take the differenceof each successive pair of results:

Then our calculations suggest that:Envelope 1 A A B B C C

Envelope 2 B C A C A B

Envelope 3 C B C A B A

-MAILORDER

Page 2: Australian Mathematics Teacher vol. 65 no. 1 - ERIC · Pascal’s triangle. Each row is obtained from the row above by a simple rule. Can you see what it is? What would the next row

amt 65 (1) 2009 4

And in general,

the sign of the last term being “+” when n iseven, and “–” when n is odd.

We shall find that the two series

and

are related in a rather unexpected way.

2. (a) In the table below we sum the first eightterms of each of the above series.Complete the table, expressing each termto four decimal places.

(b) Can you spot a simple (approximate) rela-tion between your two final answers in(a)?

We write

To 12 decimal places, e = 2.718 281 828 459.

From our calculations, it also appears to betrue that

Extensions

3. The product 1.2.3 … n is often written n!(n factorial) as we have written it above.Evaluate the numbers:2! + 13! + 14! + 15! + 16! + 1Show that n! + 1 is not exactly divisible byany of the numbers 2, 3, 4, …, n. Is it aprime number? Always? (The idea wasused by Euclid to show that there is aninfinite number of prime numbers.)

4. The array of numbers below is calledPascal’s triangle. Each row is obtainedfrom the row above by a simple rule. Canyou see what it is? What would the nextrow in the triangle be?

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

Show that the numbers in the fifth roware given by

Can you express the other numbers interms of factorials?

5. See if you can construct a mnemonic tohelp you remember the first few digits ofthe number e.

Term Sum Term Sum

1.0000 1.0000

Add 2.0000 Subtract 0.0000

Add 2.5000 Add 0.5000

Add Subtract

Add Add

Add Subtract

Add Add

Add Subtract

Page 3: Australian Mathematics Teacher vol. 65 no. 1 - ERIC · Pascal’s triangle. Each row is obtained from the row above by a simple rule. Can you see what it is? What would the next row

amt 65 (1) 2009 5

Some other patterns for e

The number e is widely occurring in mathe-matics. It is of interest to find and createpatterns from different branches of the subject,because these increase our understanding.

Thus while it may be useful to know that

e = 2.718 281 828 459 045 235…

this expression tells us nothing about the struc-ture of e — how it occurs and why it isinteresting. On the other hand, we find e occur-ring naturally in some delightfully simplesequences, series and continued fractions.Some of these we have met before.

Sequences

(a)

(b)

(c)

(d)

Series

(a)

(b)

(c)

(d)

(e)

Continued fractions

(a)

(b)

(c)

We shall investigate this very special numbere further in coming issues of AMT.

Bibliography

Scott, P. R. (1974). Discovering the mysterious numbers.Cheshire.

The exponential series :http://www.ucl.ac.uk/Mathematics/geomath/level2/series/ser9.html

The Pythagoreans, who were the first to take up mathematics, not only advanced this subject but, saturated with it, they fanciedthat the principles of mathematics were the principles of all things.Aristotle, Greek philosopher (384–322 BC)