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Test: Combinatorics - 1 - Software Development MCQ


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15 Questions MCQ Test - Test: Combinatorics - 1

Test: Combinatorics - 1 for Software Development 2024 is part of Software Development preparation. The Test: Combinatorics - 1 questions and answers have been prepared according to the Software Development exam syllabus.The Test: Combinatorics - 1 MCQs are made for Software Development 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Combinatorics - 1 below.
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Test: Combinatorics - 1 - Question 1

What is combinatorics?

Detailed Solution for Test: Combinatorics - 1 - Question 1

Combinatorics is a branch of mathematics that deals with counting, arranging, and selecting objects or elements in a systematic way. It involves various techniques such as permutations, combinations, and partitions to solve problems related to counting and arrangement.

Test: Combinatorics - 1 - Question 2

How many lucky numbers are there between 1 and 100?

Detailed Solution for Test: Combinatorics - 1 - Question 2

Lucky numbers are those numbers that contain only the digits 4 and 7. Between 1 and 100, the lucky numbers are: 4, 7, 44, 47, 74, 77. There are a total of 6 lucky numbers between 1 and 100.

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Test: Combinatorics - 1 - Question 3

What is modular arithmetic?

Detailed Solution for Test: Combinatorics - 1 - Question 3

Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" when they reach a certain value called the modulus. It involves performing arithmetic operations (addition, subtraction, multiplication, and division) on numbers within the modulus.

Test: Combinatorics - 1 - Question 4

In how many ways can the letters of the word "MISSISSIPPI" be arranged?

Detailed Solution for Test: Combinatorics - 1 - Question 4

The word "MISSISSIPPI" has 11 letters, including 1 M, 4 I's, 4 S's, and 2 P's. The total number of arrangements of these letters is given by 11! / (1! * 4! * 4! * 2!) = 34,650.

Test: Combinatorics - 1 - Question 5

What is the formula for calculating the number of combinations?

Detailed Solution for Test: Combinatorics - 1 - Question 5

The number of combinations of n objects taken r at a time is given by the formula n! / (r! * (n - r)!). It represents the number of ways to choose r objects from a set of n objects without considering the order.

Test: Combinatorics - 1 - Question 6

What will be the output of the following code?
def count_lucky_numbers(n):
    count = 0
    for i in range(1, n + 1):
        if '4' in str(i) or '7' in str(i):
            count += 1
    return count

print(count_lucky_numbers(100))

Detailed Solution for Test: Combinatorics - 1 - Question 6

The code counts the number of lucky numbers between 1 and 100. A lucky number is a number that contains the digits 4 or 7. In the given code, the function 'count_lucky_numbers' iterates over the range from 1 to n (inclusive) and checks if either '4' or '7' is present in the number. If so, it increments the count. In this case, the count will be 20.

Test: Combinatorics - 1 - Question 7

What will be the output of the following code?
def pow_mod(x, n, d):
    result = 1
    for _ in range(n):
        result = (result * x) % d
    return result

print(pow_mod(3, 4, 5))

Detailed Solution for Test: Combinatorics - 1 - Question 7

The code calculates the result of pow(x, n) % d using the pow_mod function. It performs modular exponentiation, where the base x is raised to the power n, and the result is taken modulo d. In this case, 3 raised to the power 4 is 81, and 81 % 5 is 1.

Test: Combinatorics - 1 - Question 8

What will be the output of the following code?
def permutation_partitions(n, k):
    if n == 0 or k == 1:
        return 1
    elif n < k:
        return 0
    else:
        return k * permutation_partitions(n - 1, k) + permutation_partitions(n - 1, k - 1)

print(permutation_partitions(5, 3))

Detailed Solution for Test: Combinatorics - 1 - Question 8

The code calculates the number of partitions of 5 into 3 parts using the 'permutation_partitions' function. It uses a recursive approach to calculate the number of partitions. In this case, the number of partitions of 5 into 3 parts is 10.

Test: Combinatorics - 1 - Question 9

What will be the output of the following code?
def combinations(n, r):
    if r == 0 or n == r:
        return 1
    else:
        return combinations(n - 1, r - 1) + combinations(n - 1, r)

print(combinations(6, 2))

Detailed Solution for Test: Combinatorics - 1 - Question 9

The code calculates the number of 'combinations' of 6 objects taken 2 at a time using the combinations function. It uses a recursive approach to calculate the number of combinations. In this case, the number of combinations of 6 objects taken 2 at a time is 15.

Test: Combinatorics - 1 - Question 10

What is the output of the following code?
def count_permutations(n):
    if n == 0:
        return 1
    else:
        return n * count_permutations(n - 1)

def count_partitions(n, k):
    if k == 1 or n == k:
        return 1
    elif k > n:
        return 0
    else:
        return count_partitions(n - 1, k - 1) + k * count_partitions(n - 1, k)

result = count_permutations(5) % count_partitions(10, 2)
print(result)

Detailed Solution for Test: Combinatorics - 1 - Question 10

The code calculates the factorial of 4 using the 'factorial' function. The factorial of a non-negative integer n is the product of all positive integers less than or equal to n. In this case, 4! = 4 * 3 * 2 * 1 = 24.

Test: Combinatorics - 1 - Question 11

What is the output of the following code?
def count_combinations(n, r):
    if r == 0 or n == r:
        return 1
    else:
        return count_combinations(n - 1, r - 1) + count_combinations(n - 1, r)

def pow_mod(x, n, d):
    if n == 0:
        return 1
    else:
        temp = pow_mod(x, n // 2, d)
        if n % 2 == 0:
            return (temp * temp) % d
        else:
            return (x * temp * temp) % d

result = count_combinations(5, 2) % pow_mod(2, 10, 7)
print(result)

Detailed Solution for Test: Combinatorics - 1 - Question 11

The code calculates the result of count_combinations(5, 2) % pow_mod(2, 10, 7). The count_combinations function calculates the number of combinations of n objects taken r at a time, and the pow_mod function performs modular exponentiation. In this case, count_combinations(5, 2) is 10, and pow_mod(2, 10, 7) is 2. So, 10 % 2 is 0.

Test: Combinatorics - 1 - Question 12

What is the output of the following code?
def factorial(n):
    if n == 0 or n == 1:
        return 1
    else:
        return n * factorial(n - 1)

def pow_mod(x, n, d):
    if n == 0:
        return 1
    else:
        temp = pow_mod(x, n // 2, d)
        if n % 2 == 0:
            return (temp * temp) % d
        else:
            return (x * temp * temp) % d

result = factorial(5) % pow_mod(3, 4, 5)
print(result)

Detailed Solution for Test: Combinatorics - 1 - Question 12

The code calculates the result of factorial(5) % pow_mod(3, 4, 5). The factorial function calculates the factorial of a number, and the pow_mod function performs modular exponentiation. In this case, factorial(5) is 120, and pow_mod(3, 4, 5) is 1. So, 120 % 1 is 0.

Test: Combinatorics - 1 - Question 13

What is the output of the following code?
def factorial(n):
    if n == 0 or n == 1:
        return 1
    else:
        return n * factorial(n - 1)

def pow_mod(x, n, d):
    if n == 0:
        return 1
    else:
        temp = pow_mod(x, n // 2, d)
        if n % 2 == 0:
            return (temp * temp) % d
        else:
            return (x * temp * temp) % d

result = factorial(5) % pow_mod(3, 4, 5)
print(result)

Detailed Solution for Test: Combinatorics - 1 - Question 13

The code calculates the result of factorial(5) % pow_mod(3, 4, 5). The factorial function calculates the factorial of a number, and the pow_mod function performs modular exponentiation. In this case, factorial(5) is 120, and pow_mod(3, 4, 5) is 1. So, 120 % 1 is 0.

Test: Combinatorics - 1 - Question 14

What is the output of the following code?
def pow_mod(x, n, d):
    if n == 0:
        return 1
    else:
        temp = pow_mod(x, n // 2, d)
        if n % 2 == 0:
            return (temp * temp) % d
        else:
            return (x * temp * temp) % d

def count_permutations(n):
    if n == 0:
        return 1
    else:
        return n * count_permutations(n - 1)

result = pow_mod(count_permutations(5), 2, 7)
print(result)

Detailed Solution for Test: Combinatorics - 1 - Question 14

The code calculates the result of pow_mod(count_permutations(5), 2, 7). The count_permutations function calculates the number of permutations of n objects, and the pow_mod function performs modular exponentiation. In this case, count_permutations(5) is 120, and pow_mod(120, 2, 7) is 4. So, 4 % 7 is 4.

Test: Combinatorics - 1 - Question 15

What is the output of the following code?
def count_partitions(n, k):
    if k == 1 or n == k:
        return 1
    elif k > n:
        return 0
    else:
        return count_partitions(n - 1, k - 1) + k * count_partitions(n - 1, k)

def combinations(n, r):
    if r == 0 or n == r:
        return 1
    else:
        return combinations(n - 1, r - 1) + combinations(n - 1, r)

result = count_partitions(combinations(6, 2), 3)
print(result)

Detailed Solution for Test: Combinatorics - 1 - Question 15

The code calculates the result of count_partitions(combinations(6, 2), 3). The combinations function calculates the number of combinations of n objects taken r at a time, and the count_partitions function calculates the number of partitions of n into k parts. In this case, combinations(6, 2) is 15, and count_partitions(15, 3) is 10.

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