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CAT — Permutation and Combination

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Permutation and Combination
6 questions
Q1 Consider two sets A = {2, 3, 5, 7, 11, 13} and B = {1, 8, 27}. Let f be a function from A… MCQ

Consider two sets A = {2, 3, 5, 7, 11, 13} and B = {1, 8, 27}. Let f be a function from A to B such that for every element in B, there is at least one element a in A such that f(a) = b. Then, the total number of such functions f is

A.

665

B.

667

C.

537

D.

540

Correct answer: D.

540

Step 1: Total number of functions
Each of the 6 elements of A can be mapped to any of the 3 elements of
B.
Total functions = 3⁶ = 729

Step 2: Subtract functions where one element of B is not used
Choose the missing element of B in 3 ways.
The remaining 2 elements are available for all 6 elements of
A.
Number of such functions = 3 × 2⁶ = 3 × 64 = 192
Remaining functions = 729 − 192 = 537

Step 3: Add back functions where two elements of B are not used
Choose the two missing elements in 3 ways.
In each case, all 6 elements map to the only remaining element.
Number of such functions = 3 × 1 = 3

Hence, required number of onto functions = 537 + 3 = 540

Q2 − The sum of all four-digit numbers that can be formed with the distinct non-zero digits … TITA

− The sum of all four-digit numbers that can be formed with the distinct non-zero digits a, b, c, and d, with each digit appearing exactly once in every number, is 153310 + n, where n is a single digit natural number. Then, the value of (a + b + c + d + n) is

Answer: 31

Step 1: Find the expression for the total sum
There are 4! = 24 numbers.
Each digit appears 24/4 = 6 times in each place value.
Total sum = 6 × (1000 + 100 + 10 + 1) × (a + b + c + d)
= 6 × 1111 × (a + b + c + d)
= 6666(a + b + c + d)

Step 2: Find n
6666(a + b + c + d) = 153310 + n
6666 × 23 = 153318
153310 + n = 153318 → n = 8

Step 3: Find the sum of the digits
a + b + c + d = 153318 / 6666 = 23
a + b + c + d + n = 23 + 8 = 31

Q3 A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer… MCQ

A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer can choose one of 4 breads and opt for either small or large sized sandwich. Optionally, the customer may also add up to 2 out of 6 available sauces. The number of different ways in which an order can be placed for a sandwich, is

A.

880

B.

840

C.

800

D.

600

Correct answer: A.

880

Base choices: 5 sandwich types × 4 breads × 2 sizes = 40 ways.

Sauce choices (up to 2 from 6): no sauce = 1, exactly 1 = ⁶C₁ = 6, exactly 2 = ⁶C₂ = 15. Total sauce choices = 22.

Total orders = 40 × 22 = 880.

Q4 − P, Q, R and S are four towns. One can travel between P and Q along 3 direct paths, betw… TITA

− P, Q, R and S are four towns. One can travel between P and Q along 3 direct paths, between Q and S along 4 direct paths, and between P and R along 4 direct paths. There is no direct path between P and S, while there are few direct paths between Q and R, and between R and S. One can travel from P to S either via Q, or via R, or via Q followed by R, respectively, in exactly 62 possible ways. One can also travel from Q to R either directly, or via P, or via S, in exactly 27 possible ways. Then, the number of direct paths between Q and R is

Answer: 7

Given:
● Direct paths between P and Q = 3
● Direct paths between Q and S = 4
● Direct paths between P and R = 4
● No direct path between P and S
● Let the number of direct paths between Q and R = x
● Let the number of direct paths between R and S = y
Also,
● Number of ways to travel from P to S is 62.
● Number of ways to travel from Q to R is 27.
Find x.

Step 1: Form the equation for travel from P to S
Travel is possible in three ways.
Via Q
Ways = 3 × 4 = 12
Via R
Ways = 4 × y = 4y
Via Q and then R
Ways = 3 × x × y = 3xy
Hence,
12 + 4y + 3xy = 62
or,
4y + 3xy = 50
y(4 + 3x) = 50

Step 2: Form the equation for travel from Q to R
Travel is possible in three ways.
Directly
Ways = x
Via P
Ways = 3 × 4 = 12
Via S
Ways = 4 × y = 4y
Hence,
x + 12 + 4y = 27
or,
x + 4y = 15

Step 3: Solve the equations
From the second equation,
4y = 15 − x
Substitute into the first equation,
((15 − x)/4)(4 + 3x) = 50
Multiply by 4,
(15 − x)(4 + 3x) = 200
Expand,
60 + 45x − 4x − 3x² = 200
3x² − 41x + 140 = 0
Factorizing,
(3x − 20)(x − 7) = 0
Thus,
x = 20/3 or x = 7
Since the number of paths must be an integer,
x = 7

Final Answer
The number of direct paths between Q and R is
7

Q5 − The number of all positive integers up to 500 with non-repeating digits is TITA

− The number of all positive integers up to 500 with non-repeating digits is

Answer: 378

Given:
Find the number of positive integers up to 500 whose digits do not repeat.

Step 1: Count one-digit numbers
The positive one-digit numbers are
1 to 9.
Count = 9

Step 2: Count two-digit numbers
The tens digit can be
1 to 9
= 9 choices.
The units digit can be any digit except the tens digit.
= 9 choices.
Hence,
Total two-digit numbers
= 9 × 9
= 81

Step 3: Count three-digit numbers from 100 to 499
The hundreds digit can be
1, 2, 3 or 4
= 4 choices.
The tens digit can be any digit except the hundreds digit.
= 9 choices.
The units digit can be any digit except the first two digits.
= 8 choices.
Hence,
Total
= 4 × 9 × 8
= 288

Step 4: Count the number 500
The number 500 has repeated digit 0.
Hence, it is not counted.

Step 5: Find the total
Total numbers
= 9 + 81 + 288
= 378

Final Answer
378

Q6 The number of all natural numbers up to 1000 with non-repeating digits is MCQ

The number of all natural numbers up to 1000 with non-repeating digits is

A.

504

B.

648

C.

738

D.

585

Correct answer: C.

738

We count the natural numbers from 1 to 1000 whose digits do not repeat.
1-digit numbers: These are 1 to 9. Count = 9
2-digit numbers: The first digit can be chosen in 9 ways (1 to 9). The second digit can be chosen in 9 ways (0 to 9 except the first digit). Total = 9 × 9 = 81
3-digit numbers: The first digit can be chosen in 9 ways (1 to 9). The second digit can be chosen in 9 ways (all digits except the first). The third digit can be chosen in 8 ways (excluding the first two digits). Total = 9 × 9 × 8 = 648
Number 1000: 1000 has repeated digit 0. Hence, it is not counted.
Total count = 9 + 81 + 648 = 738
Answer:
C. 738

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