536 Puzzles and Curious Problems

Answers 381

in 3 ways, two in 3 ways, and three in I way, making together 7 ways, which
is the same as the cube of 2 less l.

446. A GENERAL ELECTION

The answer is 39,147,416 different ways. Add 3 to the number of members

(making 618) and deduct I from the number of parties (making 3). Then the

answer will be the number of ways in which 3 things may be selected from 618.

That is:

618 X 617 X 616 _ 39147416

I X 2 X 3 - , , ways.

The general solution is as follows. Let p = parties and m = members. The

number of ways the parliament can be elected is equal to the number of com-

binations in which p - I objects may be selected from m + p - I objects.

447. THE MAGISTERIAL BENCH

Apart from any conditions, ten men can be arranged in line in IO! ways =

3,628,800. Now how many of these cases are barred? Regard two of a nation-

ality in brackets as one item.

(I) Then (EE) (SS) (WW) FISA can be permuted in 7! X 2^3 ways =

40,320. Remember the two E's can change places within their bracket wherever

placed, and so with the S's and the W's. Hence the 23.

(2) But we may get (EE) (SS) WWFISA, where the W's are not bracketed,

but free. This gives 8! X 22 cases, but we must deduct result (I) or these will

be included a second time. Result, 120,960.

(3) Deal similarly with the two S's unbracketed. Result, 120,960.

(4) Deal again, with the E's unbracketed. Result, 120,960.

(5) But we may have (EE) SSWWFISA, where both Sand Ware un-

bracketed. This gives 9! X 2 cases, but we must deduct results (I), (2), and

(3) for reasons that will now be obvious. Result, 443,520.

(6) When only S is bracketed, deducting (I), (2), and (4). Result, 443,520.

(7) When only W is bracketed, deducting (I), (3), and (4). Result,

443,520.