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Venn diagrams

Solution. From (1.10), the total number of distinct bridge dealings is 52!/(13!13!13!13!).
However, the number of ways in which the four aces can be distributed with one in each hand
is 4!/(1!1!1!1!) = 4!; the remaining 48 cards can then be dealt out in 48!/(12!12!12!12!) ways.
Thus the probability that each player receives an ace is

48! (13!) 4 24(13) 4
4! = = 0.105.
(12!) 4 52! 49 · 50 · 51 · 52

Asinthecaseofpermutationswemightaskhowmanycombinationsofk objectscanbechosen
from n with replacement (repetition). To compute this, we may imagine the n (distinguishable)
objects set out on a table. Each combination of k objects can then be made by pointing to k of the
n objects in turn (with repetitions allowed). These k equivalent selections distributed amongst n
differentbutre-choosableobjectsarestrictlyanalogoustotheplacingofk indistinguishable’balls’
in n different boxes with no restriction on the number of balls in each box. A particular selection
in the case k = 7, n = 5 may be symbolised as

xxx| |x|xx|x.

This denotes three balls in the first box, none in the second, one in the third, two in the fourth and
one in the fifth. We therefore need only to consider the number of (distinguishable) ways in which
k crosses and n − 1 vertical lines can be arranged, i.e. the number of permutations of k + n − 1
objects of which k are identical crosses and n − 1 are identical lines. This is given by (1.7) as

(k + n − 1)! k
= C k = C . (1.11)
k!(n − 1)! n+k−1 n

We note that this expression also occurs in the binomial expansion for negative integer powers. If
n is a positive integer, it is straightforward to show that

∞
∑
k
b ,
(a + b) −n = (−1) C k a −n−k k
n+k−1
k=0
where a is taken to be larger than b.
2
3
Example 1.10. M = {1, 2, 3}; n = 3, k = 2; C = C = 4! = 6. Indeed, there are such
2 4 2!2!
possible combinations: {1; 2}, {1; 3}, {2; 3}, {1; 1}, {2; 2}, {3; 3}. It is only six. ,

Venn diagrams

We call a single performance of an experiment a trial and each possible result an outcome. The
sample space S of the experiment is then the set of all possible outcomes of an individual trial.
For example, if we throw a six-sided die then there are six possible outcomes that together form
the sample space of the experiment. At this stage we are not concerned with how likely a
particular outcome might be (we will return to the probability of an outcome in due course) but
rather will concentrate on the classification of possible outcomes. It is clear that some sample
spaces are finite (e.g. the outcomes of throwing a die) whilst others are infinite (e.g. the
outcomes of measuring people’s heights). Most often, one is not interested in individual
outcomes but in whether an outcome belongs to a given subset A (say) of the sample space S;
these subsets are called events. For example, we might be interested in whether a person is
taller or shorter than 180 cm, in which case we divide the sample space into just two events:
namely, that the outcome (height measured) is (i) greater than 180 cm or (ii) less than 180 cm.

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