Computers and Technology, 09.10.2019 21:00 Sk8terkaylee

Aperson gets 13 cards of a deck. let us call for simplicity the types of cards by 1,2,3,4. in how many ways can we choose 13 cards, so that so that there are 5 of type 1, there are 4 of type 2, there are 2 of type 3 and there are 2 of type 4?

Answers: 2

Show answers

Answers

There are 5,598,527,220 ways to choose 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 cards of type 4 from a set of 13 cards.

Explanation:

The crucial point of this problem is to understand the possible ways of choosing any type of card from the 13-card deck.

This is a problem of combination since the order of choosing them does not matter here, that is, the important fact is the number of cards of type 1, 2, 3 or 4 we can get, no matter the order that they appear after choosing them.

So, the question for each type of card that we need to answer here is, how many ways are there of choosing 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 are of type 4 from the deck of 13 cards?

The mathematical formula for combinations is $\\ \frac{n!}{(n-k)!k!}$ , where n is the total of elements available and k is the size of a selection of k elements from which we can choose from the total n.

Then,

Choosing 5 cards of type 1 from a 13-card deck:

$\frac{n!}{(n-k)!k!} = \frac{13!}{(13-5)!5!} = \frac{13*12*11*10*9*8!}{8!*5!} = \frac{13*12*11*10*9}{5*4*3*2*1} = 1,287$ , since $\\ \frac{8!}{8!} = 1$ .

Choosing 4 cards of type 2 from a 13-card deck:

$\\ \frac{n!}{(n-k)!k!} = \frac{13!}{(13-4)!4!} = \frac{13*12*11*10*9!}{9!4!} = \frac{13*12*11*10}{4!}= 715$ , since $\\ \frac{9!}{9!} = 1$ .

Choosing 2 cards of type 3 from a 13-card deck:

$\\ \frac{n!}{(n-k)!k!} =\frac{13!}{(13-2)!2!} = \frac{13*12*11!}{11!2!} = \frac{13*12}{2!} = 78$ , since $\\ \frac{11!}{11!}=1$ .

Choosing 2 cards of type 4 from a 13-card deck:

It is the same answer of the previous result, since

$\\ \frac{n!}{(n-k)!k!} = \frac{13!}{(13-2)!2!} = 78$ .

We still need to make use of the Multiplication Principle to get the final result, that is, the ways of having 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 cards of type 4 is the multiplication of each case already obtained.

So, the answer about how many ways can we choose 13 cards so that there are 5 of type 1, there are 4 of type 2, there are 2 of type 3 and there are 2 of type 4 is:

1287 * 715 * 78 * 78 = 5,598,527,220 ways of doing that (or almost 6 thousand million ways).

In other words, there are 1287 ways of choosing 5 cards of type 1 from a set of 13 cards, 715 ways of choosing 4 cards of type 2 from a set of 13 cards and 78 ways of choosing 2 cards of type 3 and 2 cards of type 4, respectively, but having all these events at once is the multiplication of all them.

Answer from: Quest

8 track im pretty sure came first

Answer from: Quest

a software system that is used for viewing and creating hypertext is a hypertext system.

i hope this .

Another question on Computers and Technology

Computers and Technology, 22.06.2019 17:00

The two main ways in which marketers address the competition with their strategies are by satisfying a need better than a competition and by

Answers: 2

Answer

Computers and Technology, 22.06.2019 22:30

Alex’s family members live in different parts of the world. they would like to discuss the wedding plans of one of their distant relatives. however, alex wants all the family members to talk to each other simultaneously so that they can make decisions quickly. which mode of internet communication should they use? a. blog b. email c. wiki d. message board e. instant messaging

Answers: 2

Answer

Computers and Technology, 22.06.2019 22:40

Write a program that defines symbolic names for several string literals (chars between quotes). * use each symbolic name in a variable definition. * use of symbolic to compose the assembly code instruction set can perform vara = (vara - varb) + (varc - vard); ensure that variable is in unsigned integer data type. * you should also further enhance your symbolic logic block to to perform expression by introducing addition substitution rule. vara = (vara+varb) - (varc+vard). required: debug the disassembly code and note down the address and memory information.

Answers: 3

Answer

Computers and Technology, 23.06.2019 18:30

Janice recently received her college degree and is looking for a job. she is worried that since she just finished school, she will be required to repay her perkins and direct subsidized loans immediately. janice pulls out the paperwork she signed and reviews it again for repayment information. after reading all of the information, janice discovers that

Answers: 2

Answer

You know the right answer?

Aperson gets 13 cards of a deck. let us call for simplicity the types of cards by 1,2,3,4. in how ma...

Questions