Permutations and Combinations Test 16

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Permutations and Combinations Test 16

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Weekly Quiz Competition

Question 1
1 / -0

A vehicle registration number consists of $$2$$ letters of English alphabet followed by $$4$$ digits, where the first digit is not zero. Then, the total number of vehicles with distinct registration numbers is

A
$$26^{2} \times 10^{4}$$

B
$$^{26}P_{2} \times\ ^{10}P_{4}$$

C
$$^{26}P_{2} \times 9\times\ ^{10}P_{3}$$

D
$$26^{2} \times 9\times 10^{3}$$

Solution

The total number of arrangements of $$2$$ letters of English alphabet
$$= 26\times 26$$
The total number of arrangements of $$4$$ digits number in which first digit is not zero
$$= 9\times 10\times 10\times 10$$
$$\therefore$$ The total number of vehicles with distinct registration number
$$= 26\times 26\times 9\times 10\times 10\times 10$$
$$= 26^{2} \times 9\times 10^{3}$$
Question 2
1 / -0

If $$\displaystyle ^{ 16 }{ C }_{ r }=^{ 16 }{ C }_{ r+1 }$$, then the value of $$\displaystyle ^{ r }{ P }_{ r-3 }$$ is
$$\displaystyle $$

A
31

B
120

C
210

D
840

E
No option is correct

Solution

$$\displaystyle ^{ 16 }{ C }_{ r }=^{ 16 }{ C }_{ r+1 }$$

$$\displaystyle \Rightarrow \quad ^{ 16 }{ C }_{ 16-r }=^{ 16 }{ C }_{ r+1 }\quad \left( \because \quad ^{ n }{ C }_{ n-r } \right) $$

$$\displaystyle \Rightarrow \quad 16-r=r+1$$
$$\displaystyle \Rightarrow \quad 2r=15$$

$$\displaystyle \Rightarrow \quad r=7.5$$

Which is not possible, since r should be an integer.
Question 3
1 / -0

If $$X, Y, Z, D, E$$ and $$F$$ are $$6$$ distinct points on the circumference of a circle, find the number of different chords which can be drawn using any $$2$$ of the $$6$$ points.

A
$$6$$

B
$$12$$

C
$$15$$

D
$$30$$

E
$$36$$

Solution

To draw a chord we choose any two points from given 6 points .
So
The answer required is $$^6C_2=15$$
Question 4
1 / -0

If $$_{ }^{ n }{ { P }_{ r } }=30240$$ and $$_{ }^{ n }{ { C }_{ r } }=252$$, then the ordered pair $$\left( n,r \right) $$ is equal to

A
$$\left( 12,6 \right) $$

B
$$\left( 10,5 \right) $$

C
$$\left( 9,4 \right) $$

D
$$\left( 16,7 \right) $$

Solution

Given that, $$_{ }^{ n }{ { P }_{ r } }=30240$$ and $$_{ }^{ n }{ { C }_{ r } }=252$$
$$\Rightarrow \dfrac { n! }{ \left( n-r \right) ! } =30240$$ and $$\dfrac { n! }{ \left( n-r \right) !r! } =252$$
$$\Rightarrow r!=\dfrac { 30240 }{ 252 } =120$$
$$\Rightarrow r=5$$
$$\therefore \dfrac { n! }{ \left( n-5 \right) ! } =30240$$
$$\Rightarrow n\left( n-1 \right) \left( n-2 \right) \left( n-3 \right) \left( n-4 \right) =30240$$
$$\Rightarrow n\left( n-1 \right) \left( n-2 \right) \left( n-3 \right) \left( n-4 \right) =10\left( 10-1 \right) \left( 10-2 \right) \left( 10-3 \right) \left( 10-4 \right) $$
$$\Rightarrow n=10$$
Hence, required ordered pair is $$\left( 10,5 \right) $$
Question 5
1 / -0

The value of $$\dfrac {(n + 2)! - (n + 1)!}{n!} $$ is:

A
$$(n + 2)!$$

B
$$(n + 1)!$$

C
$$(n + 2)^{2}$$

D
$$(n + 1)^{2}$$

E
$$n$$

Solution

The value of $$\dfrac { (n+2)!-(n+1)! }{ n! } $$
$$=\dfrac { (n+2)(n+1)n!-(n+1)n! }{ n! } $$
$$=\dfrac { (n+1)(n!)(n+2-1) }{ n! } $$
$$=\dfrac { (n+1)(n+1) }{ 1 } $$
$$={ (n+1) }^{ 2 }$$
Question 6
1 / -0

If $$^{n}C_{12} = ^{n}C_{6}$$, then $$^{n}C_{2}$$is equal to:

A
$$72$$

B
$$153$$

C
$$306$$

D
$$2556$$

Solution

Given $$^{n}C_{12} = ^{n}C_{6}$$
or $$^{n}C_{n - 12} = ^{n}C_{6}$$
$$\Rightarrow n - 12 = 6$$
$$\Rightarrow n = 18$$
$$\therefore ^{n}C_{2} = ^{18}C_{2} = \dfrac {18\times 17}{2\times 1}$$
$$= 153$$
Question 7
1 / -0

$$^n C_{r-1}\, =\, 330,\, ^nC_r\, =\, 462,\, ^nC_{r+1}\, =\, 462\, \Rightarrow\, r\, =$$

A
3

B
4

C
5

D
6

Solution

$$^nC_{r-1}=330$$, $$^nC_r=^nC_{r+1}=462$$

$$\Rightarrow \dfrac{n!}{(r!)(\,n-r)!}=\dfrac{n!}{(r+1)!( \, n-(r+1))!}$$

$$\Rightarrow (r+1)!(n-(r+1))!=(r!)\,(n-r)!$$

$$\Rightarrow r+1=n-r$$

$$n=2r+1$$

$$\dfrac{n-r+1}{r}=\dfrac{462}{330}$$

$$= 330n-330r+330=462r$$

$$=330n+330=792r$$

$$=(n+1)330=792r$$

$$=(2r+2)330=792r$$

$$=660r+660=792r$$

$$=660=132r$$

$$\Rightarrow r=\dfrac{660}{132}$$

$$\rightarrow r=5$$
Question 8
1 / -0

If $$a=99^{50}+100^{50}$$ and $$b=101^{50}$$, then :

A
$$ a < b$$

B
$$a=b$$

C
$$a > b$$

D
$$a-b=100^{49}$$

Solution

$$a = 99^{50} + 100^{50}$$
$$b = 101^{50} = (100 + 1)^{50}$$
$$b = ^{100} C_{0} 100^{50} + ^{100}C_{1} 100^{49} + ... + ^{100}C_{100} 100^{0} = 2 \times {100}^{50} + ^{100}C_2 100^{48} + .... + ^{100}C_{100} 100^{0}$$

$$b = 100^{50} + 100^{50} + ^{100}C_2 100^{48} + .... + ^{100}C_{100} 100^{0}$$
$$b = a - 99^{50} +100^{50} + ^{100}C_2 100^{48} + .... + ^{100}C_{100} 100^{0}$$
$$\because \left(100^{50} - 99^{50}\right) > 1$$ and $$^{100}C_2 100^{48} + .... + ^{100}C_{100} 100^{0} >1$$

Hence, $$b > a $$
Question 9
1 / -0

If n is an integer with $$0\le n \le 11$$ then the minimum value of $$n!(11-n)!$$ is attained when a value of n =

A
11

B
5

C
7

D
9

Solution

Let $$y=n!(11-n)!$$
Consider $$x= ^{11}C_n=\dfrac{11!}{n!(11-n)!}$$
For maximum value of $$x$$ we must have $$n=6$$ or $$n=5$$
Thus $$x_{max}=^{11}C_6=\dfrac{11!}{5!6!}=\dfrac{11!}{y_{min}}$$
$$\Rightarrow y_{min}=5!6!$$ i.e. $$n=6$$ or $$n=5$$
Question 10
1 / -0

The number of words that can be formed out of the letters of the word "ARTICLE" so that the vowels occupy even places is

A
$$574$$

B
$$36$$

C
$$754$$

D
$$144$$

Solution

There are total $$7$$ words then there will be $$7$$ positions out of $$3$$ are even place like $$2,4,6$$ and we have $$3$$ vowels $$A,I,E$$

The vowels can be arranged in the even place in $$3!$$ places and the rest can be arranged in $$4!$$

$$\therefore$$ Required number of ways $$3!\times4!=144$$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 16

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Permutations and Combinations Test 16

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SHARING IS CARING

Question 1

$$26^{2} \times 10^{4}$$

$$^{26}P_{2} \times\ ^{10}P_{4}$$

$$^{26}P_{2} \times 9\times\ ^{10}P_{3}$$

$$26^{2} \times 9\times 10^{3}$$

Solution

Question 2

31

120

210

840

No option is correct

Solution

Question 3

$$6$$

$$12$$

$$15$$

$$30$$

$$36$$

Solution

Question 4

$$\left( 12,6 \right) $$

$$\left( 10,5 \right) $$

$$\left( 9,4 \right) $$

$$\left( 16,7 \right) $$

Solution

Question 5

$$(n + 2)!$$

$$(n + 1)!$$

$$(n + 2)^{2}$$

$$(n + 1)^{2}$$

$$n$$

Solution

Question 6

$$72$$

$$153$$

$$306$$

$$2556$$

Solution

Question 7

3

4

5

6

Solution

Question 8

$$ a < b$$

$$a=b$$

$$a > b$$

$$a-b=100^{49}$$

Solution

Question 9

11

5

7

9

Solution

Question 10

$$574$$

$$36$$

$$754$$

$$144$$

Solution

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