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,

\(B = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{3^2} \right\rfloor + \left\lfloor \frac{50}{3^3} \right\rfloor + \left\lfloor \frac{50}{3^4} \right\rfloor\)

\(B = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{9} \right\rfloor + \left\lfloor \frac{50}{27} \right\rfloor + \left\lfloor \frac{50}{81} \right\rfloor\)

\(B = 16 + 5 + 1 + 0 = 22\)

∴ 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,

\({{\frac{(2 n) !}{3 !(2 n-3) !}}\over{\frac{n !}{2 !(n-2) !}}}={44\over 3}\)

⇒ \({{\frac{(2 n) (2n-1)(2n-2)(2n-3)!}{3 .2.1.(2 n-3) !}}\over{\frac{n(n-1)(n-2) !}{2.1.(n-2) !}}}={44\over 3}\)

⇒ \({{\frac{(2 n) (2n-1)(2n-2)}{3 .2.1}}\over{\frac{n(n-1)}{2.1}}}={44\over 3}\)

⇒ \({\frac{2n(2n-1).2(n-1)}{6}\times{2\over n(n-1)}}={44\over 3}\)

⇒ 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

⇒ \(\frac{(n+1)!}{(r+1)!(n-r)!}×\frac{(n-r)!r!}{n!}= \frac{11}{6}\) 

⇒ \(\frac{n+1}{r+1}=\frac{11}{6}\)

⇒ 6n + 6 = 11r + 11

⇒ 6n = 11r + 5 ...(i)

Now, nCr ∶ n-1Cr-1 = 6 ∶ 3

⇒ \(\frac{n!}{r!(n-r)!}×\frac{(n-r)!(r-1)!}{(n-1)!}= \frac{6}{3}\)

⇒ \(\frac{n}{r}=\frac{6}{3}\)

⇒ 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) = \(\rm \frac{n!}{r!(n-r)!}\)
• 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 × \(\rm \frac{n!}{5!(n-5)!}\) = \(\rm \frac{n!}{4!(n-4)!}\) + \(\rm \frac{n!}{6!(n-6)!}\)

⇒ \(\rm \frac{1}{60(n-5)(n - 6)!} = \frac{1}{24(n-4)(n - 5)(n - 6)!} + \frac{1}{720(n - 6)!}\)

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

⇒ \(\rm \frac{1}{5} - \frac{1}{2(n-4)} = \frac{n - 5}{60}\)

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: \(\frac{1}{{5!}} + \frac{1}{{6!}} = \frac{x}{{7!}}\)

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

⇒ \(\frac{1}{{5!}} + \frac{1}{{6\; \times \;5!}} = \frac{x}{{7 \;\times \;6 \;\times \;5!}}\)

⇒ \(\frac{1}{{5!}} \times \left( {1 + \frac{1}{{6}}} \right) = \frac{1}{{5!}} \times \frac{x}{{42}}\)

⇒ \(1 + \frac{1}{{6}} = \frac{x}{{42}}\)

⇒ 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 = \(\rm \frac{n!}{r! (n-r)!}\)

nCr = nCn- r

Calculation

As we know nCr = \(\rm \frac{n!}{r! (n-r)!}\)

So, nCn = \(\rm \frac{n!}{n! (n-n)!} = \frac {1}{0!} = 1\)

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!.

\(\rm \therefore \dfrac{n!(n-1)!}{(n+1)!}=\dfrac{n!\times (n-1)!}{(n+1)\times n!}=\dfrac{(n-1)!}{n+1}\).

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