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Chapter 1: Problem 25

The game of bridge is played by 4 players, each of whom is dealt 13 cards. How many bridge deals are possible?

Short Answer

Expert verified
There are a total of 1431 possible bridge deals. This was calculated using the concept of stars and bars and dividing the 52 cards into 4 groups of 13 cards for each player (\( \binom{54}{2} = 1431\)).

Step by step solution

01

Divide the cards among the first 3 players

First, we need to find out how many cards can be divided among the first 3 players. We can use the stars and bars method. We have 52 cards (stars) and we need to divide them into 3 groups, implying we need two bars to separate them. So, for example, we could have (13 cards for player 1 | 13 cards for player 2 | 26 cards for player 3). This problem can be represented as finding the number of ways of arranging 52 stars and 2 bars. The total number of objects is 52+2=54; we need to choose the positions for the 2 bars. There are \( \binom{54}{2}\) ways to choose the positions for the 2 bars among the 54 objects.
02

Find the remaining cards for the fourth player

After dividing the cards among the first 3 players, the remaining cards are for the fourth player. If the first three players have 13 cards each, the fourth player receives the remaining 13 cards. Thus, there is only 1 way to distribute the cards for the fourth player in this case. In general, this should be a natural consequence of distributing the cards among the first 3 players. Once those are allocated, the fourth player will have a predetermined number of cards remaining. There is no choice left for the fourth player.
03

Calculate the total number of bridge deals

Since there are \( \binom{54}{2}\) ways to divide the cards among the first 3 players, and only 1 way to distribute the remaining cards to the fourth player, the total number of bridge deals is the same as the total distribution for the first three players. Therefore, the number of possible bridge deals is \( \binom{54}{2} = \frac{54!}{2!(52!)} = \frac{54 \times 53}{2} = 27 \times 53 = 1431\). So, there are a total of 1431 possible bridge deals.

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Most popular questions from this chapter

An elevator starts at the basement with 8 people (not including the elevator operator) and discharges them all by the time it reaches the top floor, number 6. In how many ways could the operator have perceived the people leaving the elevator if all people look alike to him? What if the 8 people consisted of 5 men and 3 women and the operator could tell a man from a woman?

There are \(\left(\begin{array}{l}n \\ r\end{array}\right)\) different linear arrangements of \(n\) balls of which \(r\) are black and \(n-r\) are white. Give a combinatorial explanation of this fact.

The following identity is known as Fermat's combinatorial identity. $$ \text { - }\left(\begin{array}{l} n \\ k \end{array}\right)=\sum_{i=k}^{n}\left(\begin{array}{l} i-1 \\ k-1 \end{array}\right) \quad n \geq k $$ Give a combinatorial argument (no computations are needed) to establish this identity. HINI: Consider the set of numbers 1 through \(n\). How many subsets of size \(k\) have \(i\) as their highest-numbered member?

Consider \(n\)-digit numbers where each digit is one of the 10 integers 0 , \(1, \ldots, 9 .\) How many such numbers are there for which (a) no two consecutive digits are equal; (b) 0 appears as a digit a total of \(i\) times, \(i=0, \ldots, n\) ?

From a set of \(n\) people a committee of size \(j\) is to be chosen, and from this committee a subcommittee of size \(i, i \leq j\), is also to be chosen. (a) Derive a combinatorial identity by computing, in two ways, the number of possible choices of the committee and subcommittee-first by supposing that the committee is chosen first and then the subcommittee, and second by supposing that the subcommittee is chosen first and then the remaining members of the committee are chosen. (b) Use part (a) to prove the following combinatorial identity: $$ \sum_{j=i}^{n}\left(\begin{array}{l} n \\ j \end{array}\right)\left(\begin{array}{l} j \\ i \end{array}\right)=\left(\begin{array}{l} n \\ i \end{array}\right) 2^{n-i} \quad i \leq n $$ (c) Use part (a) and Theoretical Exercise 13 to show that $$ \sum_{j=i}^{n}\left(\begin{array}{l} n \\ j \end{array}\right)\left(\begin{array}{l} j \\ i \end{array}\right)(-1)^{n-j}=0 \quad i \leq n $$
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