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BartSMP [9]
4 years ago
12

A string that contains only 0s,1s, and 2s is called a ternary string. How many strings of 10 ternary digits (0, 1, or 2) are the

re that contain exactly two 0s, three 1s, and five 2s? Generalize this to the number of ternary strings of length k with a 0’s, b 1’s, and c 2’s where a + b + c = k.
Computers and Technology
2 answers:
alexdok [17]4 years ago
8 0

Answer:

Number of strings = (10, 2)×(8,3) = 2520

Explanation:

The number of possible combinations for taking two 0's is C(10, 2)

It remains 8 Positions

The number of possible combinations for taking three 1's is C(8,3)

So there remains 5 spots

malfutka [58]4 years ago
5 0

Answer:

2520

Explanation:

There are (10 2) ways to 0,'s whereas (8 3) ways to 1's and then location of then 2's are apply.

Like

10!/2!×3!×5!.

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Write an algorithm that receives a number from the user (you can store the number in a variable called N). Then the algorithm sh
lbvjy [14]

Answer:

Algorithm:

1. Declare an integer variable N.

2. Read the value N from user.

3.While(N):

 3.1 find r=N%10;

 3.2 print r in new line.

 3.3 Update N as N=N/10.

4.end program.

Implementation in C++.

// header

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

// variable

int N;

cout<<"Enter an Integer:";

   cin>>N;

// find the digits of number

   while(N)

   {

       // last digit

       int r=N%10;

       // print last digit

       cout<<r<<endl;

       // update the number

       N=N/10;

}

return 0;

}

Output:

Enter an Integer:329                                        

9                                                          

2                                                          

3

7 0
3 years ago
4. In this problem, we consider sending real-time voice from Host A to Host B over a packet-switchednetwork (VoIP). Host A conve
mina [271]

Answer:

The time elapsed is 0.017224 s

Solution:

As per the question:

Analog signal to digital bit stream conversion by Host A =64 kbps

Byte packets obtained by Host A = 56 bytes

Rate of transmission = 2 Mbps

Propagation delay = 10 ms = 0.01 s

Now,

Considering the packets' first bit, as its transmission is only after the generation of all the bits in the packet.

Time taken to generate and convert all the bits into digital signal is given by;

t = \frac{Total\ No.\ of\ packets}{A/D\ bit\ stream\ conversion}

t = \frac{56\times 8}{64\times 10^{3}}          (Since, 1 byte = 8 bits)

t = 7 ms = 0.007 s

Time Required for transmission of the packet, t':

t' = \frac{Total\ No.\ of\ packets}{Transmission\ rate}

t' = \frac{56\times 8}{2\times 10^{6}} = 2.24\times 10^{- 4} s

Now, the time elapse between the bit creation and its decoding is given by:

t + t'  + propagation delay= 0.007 + 2.24\times 10^{- 4} s + 0.01= 0.017224 s

8 0
3 years ago
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