334 Answers to assignments
4 There are four ways of picking up the first
hat; then one has been removed, so there
are three ways of picking up the second
hat; or 4 × 3 = 12 ways of picking up the
first two hats. The total number of ways
they can pick up the four hats is 4 × 3 ×
2 × 1 = 24. We must subtract from this
the number where one person or more
has the right hat.
Look first at only one person having the
right hat. If A has the right hat, there
are 6 combinations of hat for B, C and
D (BCD, BDC, CBD, CDB, DBC, DCB).
Of these, only 2 have all BCD with the
wrong hats (CDB and DBC). The same
applies if B, C or D is the only person
with the right hat, making 8 in total.
Look now at two people having the
right hat: this could be AB, AC, AD, BC,
BD and CD. In each case, there is only
one way the other two could be wrong,
making 6 in total.
It is impossible for exactly three people
to have the right hats.
There is only one way all four people can
have the right hats.
This makes 8 + 6 + 1 = 15 ways of at
least one person having the right hat,
leaving 9 ways that everyone has the
wrong hat.You could try to list these.
6.2 Graphical methods of solution
1 Each row of tables contains 6 tables (6 ×
2 m = 12 m) with 1.5 m gaps at each end.
There must be a 1.5 m gap between the
wall and the first row of tables. Each
other row has an effective width of 0.8
+ 1.5 = 2.3 m. So the number of rows
that can fit in the room is the integer
below 13 5 23 .. = 5. Each row seats 6 × 6 + 2 =
38 people (the 2 are at the ends). 5 × 38 =
190, so A is correct.
2 The Venn diagram is as shown here. The
top-left circle represents even numbers,
the top-right circle multiples of 3 and
the bottom circle square numbers. Those
outside the three circles do not fit into
any of the categories.
10
20
28
38
8
26
34
214
22
32
15 21
33
3
27
(^612)
18 24
30
416
125
9
36
5
17 29
37
7
19
11
23
31 35
13
39
3 These statements may be represented as
a Carroll diagram.
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