## July 2009

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In my July diary I posed the following two brain-teasers:

(1)Prove that the following pattern continues:

1 + 2 = 3

1×2 + 2×3 + 3×4 = 4×5

1×2×3 + 2×3×4 + 3×4×5 + 4×5×6 = 5×6×7

…………

(2)Tell me the exact significance ofallthe following seven numbers. (Note: I have rigged it so that you almost certainly can't, but you should be able to get close).

8,549,176,320

5,289,476,310

8,315,906,742

5,289,467,310

8,290,157,346

8,290,673,451

1,367,294,850

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*Solutions*

**(1)** Several quick-witted readers spotted that we are in perms'n'coms
territory here —
permutations and combinations, that is. The key
symbol here is
*C*(*n*, *p*), the number of ways of choosing *p* from *n*. *C*(5,2) is
10: if the five objects are A, B, C, D, and E, I can
choose AB, AC, AD, AE, BC, BD, BE, CD, CE, or DE. (The order of choice doesn't matter, and you are choosing without
replacement.)

The main thing you need to know is that *C*(*n*, *p*) =
*n*! ÷ (*p*! × (*n*−*p*)!). We remember this in a
simpler form as "factorial
on the bottom, same number of terms on top." So *C*(5,2) is 5×4 divided by 2×1. It also
follows from the formula that
*C*(5,2) = *C*(5,3); of course! — 5×4 divided by 2×1 is equal to
5×4×3 divided by 3×2×1.

When you've messed with perms'n'coms a bit, this simple form becomes so ingrained that on seeing an expression like

1×2×3 + 2×3×4 + 3×4×5 + 4×5×6 = 5×6×7

you immediately want to divide through by the first tem. This reduces the whole expression to coms:

1 + *C*(4,3) + *C*(5,3) + *C*(6,3) = *C*(7,3)

Rewrite the right-hand side as *C*(7,4). How many ways can I choose 4 items from 7? If I insist on including
some particular element, there
are *C*(6,3) ways of picking the rest; if I insist on leaving that out but insist on including some other
particular one, that leaves
*C*(5,3) ways of choosing the rest; if I insist on leaving those two out but including some other one, there are
*C*(4,3) ways to pick
the rest; and if I insist on leaving out some particular three but including a fourth, I have only one way to choose
the rest. Q.E.D.

**(2)** Oh dear, I'm afraid I messed up at least one of these. Several
readers none the less got my
drift, though of course nobody nailed them all.

Write down the English words for the ten digits: zero, one, two, three, four, five, six, seven, eight, nine.
Rearrange them in alphabetic
order: eight, five, four, nine, one, seven, six, three, two, zero. That's your first number. For the others,
just do the same thing in French,
German, Italian, Russian (on the Cyrillic alphabet, of course!), Chinese (pinyin), and — just for the heck
of it —
Hungarian.

The German I know I screwed up, "hearing" *vier* with an initial "f" instead of
"v." (I've corrected it
here.) Someone tells me I got the Italian wrong, too, though I don't see it. (But he said it works for
Portuguese …)

Curious that the two closest numbers in the list are generated by Russian and Chinese, two utterly unrelated
languages.

Martin
Gardner's latest
book has some interesting facts about the first number, the one generated from English. If you divide it by five,
for example, you get
1,709,835,264, which also has each of the ten digits once! Then if you divide by five again …