Question

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

Short answer

The bridge deals with possible are$$\begin{gathered} a)26\times 26\times 10\times 10\times 10\times 10\times 10 \\ b)26\times 25\times 10\times 9\times 8\times 7\times 6 \end{gathered}$$.

Step-by-step solution

Given Information.

The game of bridge is played by $4$players, each of who is dealt $13$cards.

Part (a) Explanation.

There are $26$ways to fill the first place, and for each choice, there are$26$ways to fill the second place. That is, there are$26\times 26$ways to fill the first two places. And for each choice of those, we have $10$ways to fill the third place, and for every choice, we have $10$ways to fill the fourth place... and so on.

From the generalized basic principle of counting, there are$26\times 26\times 10\times 10\times 10\times 10\times 10$ways to generate a license plate.

Part (b) Explanation.

In the first place, there are $26$ways to fill it.

For every choice of the first place, there are $25$ways to fill the second place.

For every choice of the first two places, there are$10$ways to fill the third place.

For every choice of the first three places, there are $9$ways to fill the fourth place.

For every choice of the first four places, there are $8$ways to fill the fifth place.

For every choice of the first five places, there are $7$ways to fill the sixth place.

For every choice of the first six places, there are $6$ways to fill the seventh place.

From the generalized basic principle of counting, there are$26\times 25\times 10\times 9\times 8\times 7\times 6$ways to generate a license plate.

Explanation.

$$\begin{gathered} \left(a\right)26\times 26\times 10\times 10\times 10\times 10\times 10 \\ \left(b\right)26\times 25\times 10\times 9\times 8\times 7\times 6 \end{gathered}$$