Test: Permutations and Combinations- 1 - CA Foundation MCQ

⁴P₃

= 4!/(4!-3!)

= (4!)/1!

= 4*3*2*!

= 24

The expression ⁴P₄ refers to the number of permutations of 4 objects taken 4 at a time.

The formula for permutations is:

P(n, r) = n! / (n - r)!

For 4^P_4, n = r = 4. Therefore:

P(4, 4) = 4! / (4 - 4)! = 4! / 0!

Since 0! = 1, we have:

P(4, 4) = 4! / 1 = 4!

Now, calculate 4!:

4! = 4 x 3 x 2 x 1 = 24

Therefore, the value of ⁴P₄ is 24.

We are asked to find the value of 7! (7 factorial).

Step 1: Definition of Factorial
The factorial of a number n, denoted as n!, is the product of all positive integers from 1 to n.

Step 2: Calculating 7!
For 7!, we multiply all integers from 1 to 7:
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1

Now, let's perform the multiplication:

7 × 6 = 42
42 × 5 = 210
210 × 4 = 840
840 × 3 = 2520
2520 × 2 = 5040
5040 × 1 = 5040
Step 3: Final Answer
Therefore, the value of 7! is 5040.

0! = 1

n is a positive integer

The expression ⁿP_r refers to the number of permutations of r objects selected from a total of n objects.

The formula for permutations is:

    P(n, r) = n! / (n - r)!
    
In order for this formula to be valid, the value of n must be greater than or equal to r. This is because we cannot select more objects than we have available. Therefore, the restriction is:

n ≥ r (Option B)
So, for the permutation formula ⁿP_r to work, it is required that n ≥ r.

The expression ⁿP_r = n (n–1) (n–2) ………………(n–r–1) represents the number of permutations of r objects taken from a set of n objects.

The formula shows a sequence of factors starting from n and decreasing by 1 until we have r terms. The first factor is n, the second is n-1, and so on, until the last factor, which is n-r+1.

The total number of factors in this sequence is r, because there are r terms in the product:

• n (n–1) (n–2) ………………(n–r+1)

Therefore, the number of factors is r.

We already know that ⁿP_r can be written as 

• Recall that nPr = n! / (n-r)!. So, ⁿP₄ = n! / (n-4)! and ⁿP₂ = n! / (n-2)!
• Substituting these into the equation gives: n! / (n-4)! = 12 × (n! / (n-2)!)
• Simplifying leads to: (n(n-1))/(n-4) = 12
• This simplifies to: n(n-1) = 12(n-4)
• Expanding and solving yields n = 6.

We are given the ratio of permutations ⁿP₃ : ⁿP₂ = 3 : 1, and we need to find the value of n.

First, recall the formula for permutations:
P(n, r) = n! / (n - r)!

Using this formula, we can express the permutations:

ⁿP₃ = n! / (n - 3)! = n(n-1)(n-2)
ⁿP₂ = n! / (n - 2)! = n(n-1)
Now, we substitute these into the given ratio:
(n(n-1)(n-2)) / (n(n-1)) = 3 / 1

On simplifying:
(n-2) = 3

Solving for n:
n = 3 + 2 = 5
    
Therefore, the value of n is 5.

We are given the following two permutation equations:

^m+nP₂ = 56
^m–nP₂ = 30
First, recall the formula for permutations:

    P(n, r) = n! / (n - r)!
    
For P₂, this simplifies to:
    P(n, 2) = n(n-1)
    
Now, let's substitute the given values into the permutation formula:

^m+nP₂ = (m+n)(m+n-1) = 56
^m–nP₂ = (m-n)(m-n-1) = 30
We now have the system of equations:

    (m+n)(m+n-1) = 56
    
    (m-n)(m-n-1) = 30
    
Let's solve each equation:

1. Solve for m + n:
    (m+n)(m+n-1) = 56
    (m+n)(m+n-1) = 56
    m+n = 8 (since 8×7 = 56)
    
2. Solve for m - n:
    (m-n)(m-n-1) = 30
    (m-n)(m-n-1) = 30
    m-n = 6 (since 6×5 = 30)
    
Now, we have:
    m + n = 8
    m - n = 6
    
Adding these two equations:
    (m+n) + (m-n) = 8 + 6
    2m = 14
    m = 7
    
Substituting m = 7 into m + n = 8:
    7 + n = 8
    n = 1
    
Therefore, m = 7 and n = 1.

We are given the following two permutation equations:

• ^{n₁ + n₂}P₂ = 132
• ^{n₁ - n₂}P₂ = 30

First, recall the formula for permutations:
P(n, r) = n! / (n - r)!

For P₂, this simplifies to:
P(n, 2) = n(n - 1)

Now, let's substitute the given values into the permutation formula:

• ^{n₁ + n₂}P₂ = (n₁ + n₂)(n₁ + n₂ - 1) = 132
• ^{n₁ - n₂}P₂ = (n₁ - n₂)(n₁ - n₂ - 1) = 30

We now have the system of equations:

(n₁ + n₂)(n₁ + n₂ - 1) = 132

(n₁ - n₂)(n₁ - n₂ - 1) = 30

Let's solve each equation:

1. Solve for n₁ + n₂:

(n₁ + n₂)(n₁ + n₂ - 1) = 132 (n₁ + n₂)(n₁ + n₂ - 1) = 132 n₁ + n₂ = 12 (since 12 × 11 = 132)

2. Solve for n₁ - n₂:

(n₁ - n₂)(n₁ - n₂ - 1) = 30 (n₁ - n₂)(n₁ - n₂ - 1) = 30 n₁ - n₂ = 6 (since 6 × 5 = 30)

Now, we have:

n₁ + n₂ = 12 n₁ - n₂ = 6

Adding these two equations:

(n₁ + n₂) + (n₁ - n₂) = 12 + 6 2n₁ = 18 n1 = 9

Substituting n1 = 9 into n₁ + n₂ = 12:

9 + n₂ = 12 n₂ = 3

Therefore, n₁ = 9 and n₂ = 3.

Since all letters in the word "COMPUTER" are distinct then the arrangements is 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320

Now we have 3 consonants and 4 Vowels. 
as all vowels have to come together let treat them one object so we now have 3 consonant and 1 this entity
so this can be arranged in 4! ways and 4 vowels internally in 4! ways so 
final answer would be 4!x4! = 24x24 = 576

No. of ways in which 10 paper can arrange is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10!- 9!.2! = 9!(10-2) = 8.9!.

N. particles arrange in a way = n!
if two particles are arranged together then
there is only (n-1) particles
when two particles are together = (n-1)!
and two particles are arranged in= 2! ways
Thus, number of ways in which no particles
are not come together
=n! -2×(n-1)!
=( n-1)!×(n-2)

We are asked to find the number of ways the first, second, and third positions can be won in a quiz contest with 12 school teams.
This is a problem of permutations, since the order of positions matters.
The formula for permutations when selecting r objects from n objects is:
P(n, r) = n! / (n - r)!
In this case, we are selecting 3 teams (first, second, and third positions) from 12 teams, so n = 12 and r = 3.
We need to calculate P(12, 3):
P(12, 3) = 12! / 9! = 12 x 11 x 10    
Now, calculate the product:
12 × 11 = 132
132 × 10 = 1320
Thus, the number of ways the first, second, and third positions may be won is 1320.

We are asked to find the sum of all 4-digit numbers that can be formed using the digits 2, 4, 6, and 8 without repetition.

Step 1: Calculate the total number of 4-digit numbers
We are given four digits: 2, 4, 6, and 8, and we need to form 4-digit numbers without repetition. The total number of such numbers is simply the number of permutations of these 4 digits:
4! = 4 × 3 × 2 × 1 = 24
So, there are 24 unique 4-digit numbers.
Step 2: Contribution of each digit to the sum
Each digit (2, 4, 6, and 8) will appear in each place value (thousands, hundreds, tens, ones) the same number of times. Since there are 24 numbers and 4 places (thousands, hundreds, tens, ones), each digit will appear in each place exactly:
24 ÷ 4 = 6 times
Step 3: Calculate the total contribution of each place value
The value of a digit in the thousands place contributes 1000 times the digit, in the hundreds place 100 times the digit, in the tens place 10 times the digit, and in the ones place 1 times the digit. Therefore, the total sum is:
6 × (1000 + 100 + 10 + 1) × (2 + 4 + 6 + 8)
Step 4: Perform the calculations
First, calculate the sum of the place values:
1000 + 100 + 10 + 1 = 1111
Next, calculate the sum of the digits:
2 + 4 + 6 + 8 = 20
The total sum is:
6 × 1111 × 20 = 133320
Final Answer:
The sum of all the 4-digit numbers is 133320.

To form a 4-digit number greater than 5000 using the digits 3, 4, 5, 6, and 7 without repetition, we need to consider the following:

1. The first digit has to be either 5, 6, or 7.
2. The remaining digits can be any of the remaining digits, i.e., 4 digits out of the 4 available.

For the first digit, there are 3 choices (5, 6, or 7).

Then, for the remaining three digits, we have 4 choices for the second digit, 3 choices for the third digit, and 2 choices for the fourth digit, since we cannot repeat any digit.

So, the total number of such 4-digit numbers is 3×4×3×2 =72.

There are 4 ways to pick the 1st digit (can't use 0 for the 1st digit).
There are 4 remaining ways to pick the 2nd digit. Can pick 0, but not the digit picked for the 1st digit.
There are 3 remaining ways to pick the 3rd digit.
There are 2 remaining ways to pick the 4th digit.
Answer: (4)(4)(3)(2) = 96 ways.

angle would be 1 word  so total letters would be 4 [ Tri + angle ] these 4 letters can be arranged in 4! ie 24 ways

There are only 3 vowels, a, u, and e, and there are 4 odd places.  So
one of the odd spaces must contain a consonant.
We can choose 3 of the 4 odd places to put the vowels in C(4,3) or 4 ways.
For each of those vowel , we can  rearrange them  in P(3,3) or 3! or 6 ways.
And we can rearrange the 5 consonants in P(5,5) or 5! or 120 ways.
Hence the total number of different words formed are 6*4*120 = 2880 

We are asked to find the number of ways in which 7 girls can form a ring.

Step 1: Formula for Circular Permutation
The formula for the number of circular permutations of n objects is:
Number of circular permutations = (n - 1)!
Step 2: Apply the Formula
In this case, we have 7 girls, so n = 7. Applying the formula:
Number of ways to arrange 7 girls in a ring = (7 - 1)! = 6!
Step 3: Calculate 6!
Now, calculate the factorial of 6:
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
Final Answer:
The number of ways in which 7 girls can form a ring is 720.

No. of ways in which n people can sit in a round table = (n-1)! 
So to make them sit together, consider them as a single unit in the sitting arrangement ie. n= 6.
Number of ways so that two particular boys may sit together in the round table:- 

(6 -1)!*2

(The two particular boys can also be switched among themselves) 

=> 120*2
=> 240 

No. of ways in which 10 paper can arranged is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!

1. Choosing the Pair of Ladies:

   • There are three ladies: Lady X, Lady Y, and Lady Z.
   • We need to choose 2 out of these 3 to sit together.
   • Number of ways to choose the pair: 3 (XY, XZ, YZ).

2. Arranging the Paired Ladies:

   • The two chosen ladies can switch places within their pair.
   • Number of arrangements for the pair: 2.

3. Selecting Seats for the Ladies:

   • After placing the gents, there are three seats left for the ladies.
   • To ensure that only two ladies sit together:
     • Choose one of the three available seats to place the pair of ladies.
     • Number of ways to choose the seat for the pair: 3.

4. Placing the Third Lady:

   • The third lady must sit in one of the remaining two seats that are not adjacent to the pair.
   • Number of ways to place her: 2.

Step 4: Calculating the Total Number of Arrangements

Now, multiply the number of ways from each step:

1. Arrangements of the remaining gents: 2
2. Choosing the pair of ladies: 3
3. Arranging the pair: 2
4. Choosing the seat for the pair: 3
5. Placing the third lady: 2

Total Ways = 2 (gents) × 3 (pairs) × 2 (arrangements) × 3 (seat choices) × 2 (placing third lady) = 72

We are asked to find the number of ways in which the letters of the word "DOGMATIC" can be arranged.

Step 1: Count the Total Number of Letters
The word "DOGMATIC" has 8 letters in total:

D, O, G, M, A, T, I, C

Step 2: Check for Repetition of Letters
In this case, all the letters in "DOGMATIC" are distinct, meaning no letters are repeated.

Step 3: Calculate the Total Number of Arrangements
Since all the letters are distinct, the total number of ways to arrange 8 distinct letters is given by the factorial of the number of letters:

8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320

Final Answer:
The number of ways in which the letters of the word "DOGMATIC" can be arranged is 40320.

Number of different things = 10

Numbers of things which are taken at a time = 4

Now, it is said that one particular thing always occur.

So The question reduces to selecting 9 things from the 3 things

This can be done easily by using the concept of combinations

Number of ways selecting 3 things out of 9 different things such that one particular thing always occur = 84

Now, the 4th thing is fixed so number of ways for selecting the 4th thing = 4! = 24

So, Required number of ways = 84 × 24 = 2016