Factorial and its Properties MCQ Quiz in తెలుగు - Objective Question with Answer for Factorial and its Properties - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Some useful formulas are:

n! = n.(n - 1).(n - 2)...3.2.1

Calculation:

Given expression is \(\rm {1\over 9!}+{1\over 8!}={x\over 10!}\)

∴ \(\rm {1\over 9\;\times\; 8!}+{1\over 8!}={x\over 10\;\times \;9\;\times\; 8!}\)

∴ \(\rm {1\over 8!}({{1\over 9} +1})={x\over 10\;\times\; 9\;\times \;8!}\)

∴ \({10\over 9} ={ x\over 10\;\times\; 9}\)

∴ x = 100

\(\frac{{(x + 2)!}}{{(2x - 1)!}} \times \frac{{(2x + 1)!}}{{(x + 3)!}} = \frac{{72}}{7}\)

\(\frac{{(x + 2)!}}{{(2x - 1)!}} \times \)\(\frac{{(2x + 1)(2x)(2x - 1)!}}{{(x + 3)(x + 2)!}} = \frac{{72}}{7}\)

\( = \frac{{2x(2x + 1)}}{{x + 3}} = \frac{{72}}{7}\)

\( = \frac{{x(2x + 1)}}{{x + 3}} = \frac{{36}}{7}\)

= 14x² + 7x = 36x + 108

= 14x² - 29x - 108 = 0

\(x = \frac{{ - ( - 29) \pm \sqrt {{{( - 29)}^2} + 4 \times 14 \times 108} }}{{2 \times 14}}\)

\( = \frac{{29 \pm 83}}{{28}}\)

\(x = \frac{{29 + 83}}{{28}}\), \(x = \frac{{29 - 83}}{{28}}\)

x = 4, -1.92

Given x ϵ N, N → Natural number 

x = 4

Concept:

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

Calculation:

We have:

(n + 1)! = 6 × (n - 1)!

⇒ (n + 1) × n × (n - 1)! = 6 × (n - 1)!

⇒ (n + 1) × n = 6

⇒ n2 + n - 6 = 0n

⇒ n2 + 3n - 2n - 6 = 0

⇒ n(n + 3) - 2(n + 3) = 0

⇒ (n + 3)(n - 2) = 0

⇒ n + 3 = 0 OR n - 2 = 0

⇒ n = - 3 OR n = 2.

Since, n has to be a natural number, n = 2.

Concept:

nPr = \(n! \over (n-r)!\),

 nCr = \(n! \over r!(n-r)!\)

Calculation:

Given,  nPr = 840, ⇒ \(n! \over (n-r)!\) = 840  ___(i)

nCr = 35,  ⇒ \(n! \over r!(n-r)!\) = 35   ___(ii)

Concept:

C(n, r) = \(\rm\dfrac{n!}{r!(n - r)!}\)

n! = n. (n - 1)!

Calculations:

Given, C(20, n + 2) = C(20, n - 2)

\(\rm\frac{20!}{(n+2)!(18-n)!} = \rm\frac{20!}{(n-2)!(22-n)!}\)

\(\rm\frac{1}{(n+2)!(18-n)!} = \rm\frac{1}{(n-2)!(22-n)!}\)

we know that

(n + 2)! =(n - 2)! (n - 1)n(n + 1)(n+2)

(22 - n)! = (18 - n)! (19 - n)(20 - n)(21 - n)(22 - n)

\(\rm\frac{1}{((n - 2)! (n - 1)n(n + 1)(n+2)(18-n)!} = \rm\frac{1}{(n-2)!(18 - n)! (19 - n)(20 - n)(21 - n)(22 - n))}\)

\(\rm\frac{1}{(n - 1)n(n + 1)(n+2)} = \rm\frac{1}{ (19 - n)(20 - n)(21 - n)(22 - n))}\)

\(\rm {(n - 1)n(n + 1)(n+2)} = \rm (19 - n)(20 - n)(21 - n)(22 - n)\)

Now , put n = 10 

L.H.S = (n - 1)n(n + 1)(n+2) = (9)(10)(11)(12) 

R.H.S = (19 - n)(20 - n)(21 - n)(22 - n) = (9)(10)(11)(12) 

L.H.S. = R.H.S.

Hence, if C(20, n + 2) = C(20, n - 2), then n = 10

Alternate MethodWe have,

C(20, n + 2) = C(20, n - 2)

⇒ \(^{20}C_{(n+2)}=^{20}C_{(n−2)}\)

Consider,

n = 20

x = (n + 2)

y = (n - 2)

x ≠ y,

∵ n = 20 and, n ≠ 0

x + y = n (Possible)

In that case,

(n + 2) + (n - 2) = 20

or, 2n = 20

Hence, n = 10

Concept:

Fundamental principal of multiplication:

Let us suppose there are two tasks A and B such that the task A can be done in m different ways following which the second task B can be done in n different ways. Then the number of ways to complete the task A and B in succession respectively is given by: m × n ways

Fundamental principal of addition:

Let us suppose there are two tasks A and B such that the task A can be done in m different ways and task B can be completed in n ways. Then the number of ways to complete either of the two tasks is given by: (m + n) ways.

Calculation:

Here we have to form 3 digit numbers without using the digits 0, 1, 2, 4, 5, 7 such that repetition of digits is not allowed.

The hundredth digit can be filled by any digit except  0, 1, 2, 4, 5, 7

⇒ No.of ways to fills one thousand digit = 5

⇒ No. of ways to fill hundredth digit = 5 

⇒ No. of ways to fill tenth digit = 4

⇒ No. of ways to fill unit’s digit = 3

Concept:

• \({\;^n}{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}}\)

Calculation:

Given: 42(nP2) = nP4 

As we know that, \({\;^n}{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}}\)

⇒ 42 × [n × (n - 1)] = n × (n - 1) × (n - 2) × (n - 3)

⇒ 42 = n² - 5n + 6

⇒ n² - 5n - 36 = 0

⇒ n² - 9n + 4n - 36 = 0

⇒ n × (n - 9) + 4 × (n - 9) = 0

⇒ (n + 4) × (n - 9) = 0

⇒ n = - 4 or 9

∵ n ∈ N ⇒ n = 9

Hence, option C is the correct answer.

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.

Calculation:

Given sum is
0! + 1! + 2! +.....+ 99! + 100!

As we can see, all factorials except 0!, 1!, 2!, 3! has the product of 4 & 3 in their factor so, they are completely divisible by 12. 

Therefore, we consider only on the sum of 0! + 1! + 2! + 3!

⇒ Sum = 1 + 1 + 2 + 6

⇒ Sum = 10

Which when divided by 12 leaves remainder as 10.

Concept:  

Properties of combination:

• nCr = nCn-r
• If nCx = nCy then n = x + y

Calculation:

Given that, C(28, 2r) = C(28, 2r - 4)

⇒ ²⁸C_2r = ²⁸C_2r-4

As we know, if nCx = nCy then n = x + y

Here n = 28, x = 2r and y = 2r - 4

⇒ 28 = 2r + 2r - 4

⇒ 32 = 4r

⇒ r = 8

∴ The value of r is 8