What is the output for this program?

a password to a certain database consists of digits that cannot be repeated. if the password is known to consist of at least 8 d

BaLLatris [955]

Answer:

$\frac{10!}{2}mins$

Explanation:

12 seconds to try one combination will be equivalent to $\frac{1}{12}\times 60 = \frac{1}{5} \ mins$

Password contain at least 8 digit i.e. password can contain 8, 9, 10 digit.

Password cannot contain more than 10 digit because it will give room for repetition which it is clearly stated that digit cannot be repeated.

Possible digit that can be used: 9,8,7,6,5,4,3,2,1,0.

Total number of passwords combination possible for each position in 8 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3. Total number of passwords combination possible in 8 digit is equivalent to $\frac{10!}{2}$ .

Total number of passwords combination possible for each position in 9 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3, 9th position = 2. Total number of passwords combination possible in 9 digit is equivalent to $\frac{10!}{1}$ .

Total number of passwords combination possible for each position in 10 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3, 9th position = 2, 10th position = 1. Total number of passwords combination possible in 10 digit is equivalent to 10!.

Adding them up and multiplying by $\frac{1}{5} \ mins$ to get the total number of time needed to guarantee access to database = $[\frac{10!}{2}\times\ \frac{10!}{1} \times 10!] \frac{1}{5}\ mins = \frac{10!}{2}$

7 0

4 years ago

Unanswered questionl

antoniya [11.8K]

¿ ? ¿ ? Whattttt ? ¿ ? ¿

3 0

3 years ago

Write a simple algorithm in pseudocode that asks the user their favourite colour and then agrees with their choice, quoting the

olga55 [171]

Answer:

i am still a beginner in this but i hope it helps.

add scanner object;
declare string FavColour;

print msg to ask user for input

scan input FavColour;

print msg (" your favourite colour is " + FavColour)

3 0

3 years ago

The process of finding the largest and smallest numbers is used frequently in computer applications. Write a C++ program that us

WINSTONCH [101]

Answer:

The program to this question can be given as:

Program:

#include<iostream> //header file

using namespace std;

int main() //main method

{

int x[10],i,largest = 0,second_largest=0,n; //variable

cout << "Enter Number of elements :"; //message

cin>>n;

cout << "Insert array elements :"; //message

for(i=0;i<n;i++) //insert elements in array

{

cin >>x[i];

}

//Finding Largest element

for(i=0;i<n;i++)

{

if (x[i]>largest)

{

largest = x[i];

}

//finding second largset element

for(i=0;i<n;i++)

{

if (x[i]>second_largest)

{

if(x[i]==largest)

{

continue; //Ignoring largest in order to get second largest

}

second_largest=x[i];

}

//print value

cout <<"Largest Number:"<<largest<<endl;

cout <<"Second Largest Number:"<<second_largest;

return 0;

}

Output:

Enter Number of elements :5

Insert array elements :33

45

75

87

23

Largest Number:87

Second Largest Number:75

Explanation:

In the above program firstly we define the header file then we define the main method in the main method we define the array and other variables. We first input the number for the size of the array. Then we insert array elements after inserting array elements we search the largest number and the second largest number in the array. To search the largest number in the array we use the loop. To search the first largest number we define a condition that array is greater than the largest number and store the value into the largest variable. Then we check the second largest number in the array for this we use two conditions that are array is greater than the second largest number in this we use another condition that is array is equal to the largest number. If the inner condition is true then it will move forward and end of an inner condition. In the outer condition, the value will be stored on the second_largest variable all the conditions will be done inner the loop. At the last we print values.

3 0

3 years ago

Question 5 of 50

erik [133]

Answer:

D

Explanation:

because it is more safe than a key

4 0

2 years ago