Question

A state lottery game consists of choosing one card from each of the four suits in a standard deck of playing cards. (There are 13 cards in each suit.)

Count the number of ways in which four cards, each of a different face value, can be chosen.
ways

Answer 1

Answer:

17160

Step-by-step explanation:

first there are 13 options

then there are 13 - 1 that was chosen

then there are 13 - 2 that were chosen

then there are 13 - 3 that were chosen

so that's 13 options, then 12 options, then 11 options, and then 10 options

and in order to figure out all of the possibilties we mutiply the options so

13*12*11*10 = 17160

little further explaining of why this works:

we have a set of letters (3 in a set)

A B C

what are the possible combinations?

AB

AC

BC

BA

CA

CB

the answer is 6 which is also 3 * 2 * 1 = 6

HOPE THAT HELPS!1! ^_^

Answer 2

Using the Fundamental Counting Theorem, it is found that there are 28,561 ways in which four cards, each of a different face value, can be chosen.

Fundamental counting theorem:

States that if there are n things, each with $n_1, n_2, …, n_n$ ways to be done, each thing independent of the other, the number of ways they can be done is:

$N = n_1 \times n_2 \times \cdots \times n_n$

In this problem, four cards are chosen, each with 13 ways, hence $n_1 = n_2 = n_3  = n_4 = 13$, and:

$T = 13 \times 13 \times 13 \times 13 = 13^4 = 28561$

There are 28,561 ways in which four cards, each of a different face value, can be chosen.

A similar problem is given at https://brainly.com/question/19022577