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A person 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 ca

natima [27]

Answer:

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.

5 0

3 years ago

In a bluesnooping attack, the attacker copies emails, calendars, contact lists, cell phone pictures, or videos by connecting to

Kay [80]

The answer is depends. You can only get what the port lets you get. If the phone bans sending photos over Bluetooth than you can’t get the photos

3 0

3 years ago

Which of the following statements about personality tests are true? Check all of the boxes that apply.

borishaifa [10]

Answer:

They evaluate a person’s personality traits And They can be used as self-assessment tools for career planning. are the answers.

4 0

4 years ago

Implement the RC4 stream cipher in C++. User should be able to enter any key that is 5 bytes to 32 bytes long. Be sure to discar

gladu [14]

Answer:

Explanation:

#include <iostream>

#include <string>

#include<vector>

using namespace std;

vector<int> permute(vector<int>, vector<int>);

string encrypt(vector<int>s1 , vector<int> t1, string p);

string decrypt(vector<int>s1, vector<int> t1, string p);

int main() {

string plaintext = "cryptology";

string plaintext2 = "RC4";

vector<int> S(256);

vector<int> T(256);

int key[] = { 1,2,3,6 };

int key2[] = { 5,7,8,9 };

int tmp = 0;

for (int i = 0; i < 256;i++) {

S[i] = i;

T[i] = key[( i % (sizeof(key)/sizeof(*key)) )];

}

S = permute(S, T);

for (int i = 0; i < 256 ;i++) {

cout << S[i] << " ";

if ((i + 1) % 16 == 0)

cout << endl;

}

cout << endl;

string p = encrypt(S, T, plaintext);

cout << "Message: " << plaintext << endl;

cout << "Encrypted Message: " << " " << p << endl;

cout << "Decrypted Message: " << decrypt(S, T, p) << endl << endl;

tmp = 0;

for (int i = 0; i < 256;i++) {

S[i] = i;

T[i] = key2[(i % (sizeof(key) / sizeof(*key)))];

}

S = permute(S, T);

for (int i = 0; i < 256;i++) {

cout << S[i] << " ";

if ((i + 1) % 16 == 0)

cout << endl;

}

cout << endl;

p = encrypt(S, T, plaintext2);

cout << "Message: " << plaintext2 << endl;

cout << "Encrypted Msg: " << p << endl;

cout << "Decrypted Msg: "<<decrypt(S, T, p) << endl << endl;

return 0;

}

string decrypt(vector<int>s1, vector<int> t1, string p) {

int i = 0;

int j = 0;

int tmp = 0;

int k = 0;

int b;

int c;

int * plain = new int[p.length()];

string plainT;

for (int r = 0; r < p.length(); r++) {

i = (i + 1) % 256;

j = (j + s1[i]) % 256;

b = s1[i];

s1[i] = s1[j];

s1[j] = b;

tmp = (s1[i] + s1[j]) % 256;

k = s1[tmp];

c = ((int)p[r] ^ k);

plain[r] = c;

plainT += (char)plain[r];

}

return plainT;

}

string encrypt(vector<int>s1, vector<int> t1, string p) {

int i = 0;

int j = 0;

int tmp = 0;

int k = 0;

int b;

int c;

int * cipher = new int [p.length()];

string cipherT;

cout << "Keys Generated for plaintext: ";

for (int r = 0; r < p.length(); r++) {

i = (i + 1) % 256;

j = (j + s1[i]) % 256;

b = s1[i];

s1[i] = s1[j];

s1[j] = b;

tmp = (s1[i] + s1[j]) % 256;

k = s1[tmp];

cout << k << " ";

c = ((int)p[r] ^ k);

cipher[r] = c;

cipherT += (char)cipher[r];

}

cout << endl;

return cipherT;

}

vector<int> permute(vector<int> s1, vector<int> t1) {

int j = 0;

int tmp;

for (int i = 0; i< 256; i++) {

j = (j + s1[i] + t1[i]) % 256;

tmp = s1[i];

s1[i] = s1[j];

s1[j] = tmp;

}

return s1;

}

6 0

3 years ago

Identify which of these types of sampling is used: random, systematic, convenience, stratified, or cluster. To determine her

Zarrin [17]

Answer:

B.

Explanation:

Based on the sampling methods provided in regards to the question it can be said that the method being used is called Simple Random. This refers to dividing the population into different group which are then chosen at random, giving each group an equal chance of getting chosen. Which is exactly what is going on in this scenario, as the day is divided into three parts and her mood is measured randomly during each part of the day.

6 0

4 years ago