Factorial and its Properties MCQ Quiz - Objective Question with Answer for Factorial and its Properties - Download Free PDF

Calculation:

Given,

Number, n = 50

Prime number, p = 3

The exponent of 3 in 50! is given by,

∴ The maximum value of n is 22.

Given:

Five different books (A, B, C, D and E) are to be arranged on a shelf.

Books C and D are to be arranged first and second starting from the right of the shelf.

Calculation:

Since books C and D are arranged first and second starting from the right of the self,

∴ Only books A, B and E will change order.

Thus, an arrangement problem involves 3 times and the number of different order is given by 3!

Hence, the number of different orders in which books A, B and E may be arranged is 3!.

∴ Option 2 is correct.

Explanation:

From the given condition,

⇒ 

⇒ 

⇒ 

⇒ 2n - 1 = 11

⇒ 2n = 12 ⇒ n = 6  

(2) is correct

Calculation:

(n+2) ! = 2550 × n !

(n + 2).(n + 1).n ! = 2550 n!

(n + 2).(n + 1) = 51 × 50

On comparing both sides

n + 1 = 50

∴ n = 49

Calculation:

n+1Cr+1 ∶ nCr ∶ n-1Cr-1 = 11 ∶ 6 ∶ 3

Consider, n+1Cr+1 : ⁿC_r = 11 : 6

⇒  

⇒ 

⇒ 6n + 6 = 11r + 11

⇒ 6n = 11r + 5 ...(i)

Now, nCr ∶ n-1Cr-1 = 6 ∶ 3

⇒ 

⇒ 

⇒ n = 2r ...(ii)

Substituting the value of n in (i), we get

12r = 11r + 5

⇒ r = 5

⇒ n = 2(5) = 10

∴ nr = 5×10 = 50

∴ The value of nr is 50.

CONCEPT:

n! = n × (n - 1) × ...... × 1

CALCULATION:

Here, we have to find the value of n such that (n + 2)! = 60 × (n - 1)!

As we know that, n! = n × (n - 1) × ...... × 1

⇒ (n + 2) × (n + 1) × n × (n - 1)! = 60 × (n - 1)!

⇒ (n + 2) × (n + 1) × n = 60

⇒ (n + 2) × (n + 1) × n = 5 × 4 × 3

⇒ (n + 2) × (n + 1) × n = (3 + 2) × (3 + 1) × n

⇒ n = 3

Hence, option A is the correct answer.

CONCEPT:

n! = n × (n - 1) × ...... × 1

CALCULATION:

Here, we have to find the value of n such that (n + 2)! = 2550 × n!

⇒ (n + 2) × (n + 1) × n! = 2550 × n!

⇒ n2 + 3n + 2 = 2550

⇒ n2 + 3n - 2548 = 0

⇒ n2 + 52n - 49n - 2548 = 0

⇒ n( n + 52) - 49 × (n + 52) = 0

⇒ (n + 52) × (n - 49)  = 0

⇒ n = - 52 or 49

∵ n ∈ N ⇒ n = 49

Hence, option B is the correct

CONCEPT:

n! = n × (n - 1) × ...... × 1

CALCULATION:

Here, we have to find the value of n such that (n + 2)! = 30 × n!

⇒ (n + 2) × (n + 1) × n! = 30 × n!

⇒ n² + 3n + 2 = 30

⇒ n2 + 3n - 28 = 0

⇒ n² + 7n - 4n - 28 = 0

⇒ n( n + 7) - 4(n + 7) = 0

⇒ (n - 4) × (n + 7) = 0

⇒ n = 4 or - 7

∵ n ∈ N ⇒ n = 4

Hence, option B is the correct

Formula used:

• C(n,r) = 
• If a, b and c are in A.P then 2b = a + c

Calculation:

According to the question C(n, 4), C(n, 5) and C(n, 6) are in AP

2 C(n, 5) = C(n, 4) + C(n, 6) 

⇒ 2 ×  =  + 

⇒ 

Multiply by 24(n - 5)(n - 6) on both side

⇒ 

Again, multiply 60(n - 4) on both side

⇒ 12(n - 4) = 30 + (n - 5)(n - 4)

⇒ 12n - 48 = 30 + n² - 9n + 20

⇒ n² - 21n + 98 = 0

⇒ n = 14, 7

∴ The value of n is 7.

CONCEPT:

n! = n × (n - 1) × ...... × 1

CALCULATION:

Given: 

As we know that, n! = n × (n - 1) × ...... × 1

⇒ 

⇒ 

⇒ 

⇒ x = 49

Hence, option C is the correct answer.

CONCEPT:

n! = n × (n - 1) × ...... × 1

CALCULATION:

Here, we have to find the value of n such that (n + 1)! = 12 × (n - 1)!

As we know that, n! = n × (n - 1) × ...... × 1

⇒ (n + 1) × n × (n - 1)! = 12 × (n - 1)!

⇒ n × (n + 1) = 12

⇒ n² + n - 12 = 0

⇒ n² + 4n - 3n - 12 = 0

⇒ n(n + 4) - 3(n + 4) = 0

⇒ (n - 3) × (n + 4) = 0

⇒ n =  3 or -4

∵ n ∈ N ⇒ n = 3

Hence, option B is the correct answer.

Concept:

• The factorial of a natural number n is defined as: n! = 1 × 2 × 3 × ... × n.
• 0! = 1.

Calculation:

We have:

(n + 1)! = 12 × (n – 1)!

⇒ (n + 1) × n × (n - 1)! = 12 × (n - 1)!

⇒ (n + 1) × n = 12

⇒ n² + n - 12 = 0n

⇒ n2 + 4n - 3n - 12 = 0

⇒ n(n + 4) - 3(n + 4) = 0

⇒ (n + 4)(n - 3) = 0

⇒ n + 4 = 0 OR n - 3 = 0

⇒ n = -4 OR n = 3.

Since, n has to be a natural number, n = 3.

Concept:

nC_r = 

nCr = nCn- r

Calculation

As we know nCr = 

So, nCn = 

Concept:

• The factorial (n!) is defined as the product of first n natural numbers.
• n! = n × (n - 1) × (n - 2) × … × 2 × 1.

Calculation:

We know that (n + 1)! = (n + 1) × n × (n - 1) × (n - 2) × ... × 2 × 1 = (n + 1) × n!.

.

Calculation:

(n+2) ! = 2550 × n !

(n + 2).(n + 1).n ! = 2550 n!

(n + 2).(n + 1) = 51 × 50

On comparing both sides

n + 1 = 50

∴ n = 49