Permutations and Combinations Test 45

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Permutations and Combinations Test 45

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Weekly Quiz Competition

Question 1
1 / -0

How many $$3$$ digit numbers can we make using the digits $$4,5,7$$ and $$9$$ and where repetition is allowed?

A
$${3}^{4}$$

B
$${3}^{3}$$

C
$${4}^{4}$$

D
$$12$$

E
$${4}^{3}$$

Solution

We have four digits.
The numbers we need to make have $$3$$ digits.
There are $$4$$ choices for the first digit, $$4$$ choices for the second digit and $$4$$ choices for the third digit since repetition is allowed.
Hence, the total number of $$3$$ digit numbers that can be made is equal to:
$$4\times 4\times 4={ 4 }^{ 3 }$$

Hence, option E is correct.
Question 2
1 / -0

A password for a computer system requires exactly $$6$$ characters. Each character can be either one of the $$26$$ letters from A to Z or one of the ten digits from $$0$$ to $$9$$. The first character must be a letter and the last character must be a digit. How many different possible passwords are there?

A
Less than $$10^{7}$$

B
Between $$10^{7}$$ and $$10^{8}$$

C
Between $$10^{8}$$ and $$10^{9}$$

D
Between $$10^{9}$$ and $$10^{10}$$

E
More than $$10^{10}$$

Solution

The password required has to be of 6 characters and the first of those should be one of the 26 letters, the last must be one of the 10 digits and the remaining four can be either of those 36.
Thus, the possibilities for first place become 26, for the next four places become 36, and for the last place become 10.
The total number of possibilities therefore become $$36 \times (26)^4 \times 10 = 16,45,11,360$$
This can be written as $$1.6451136 \times 10^8$$ and thus lies between $$10^8$$ and $$10^9$$
Question 3
1 / -0

The exponent of $$5$$ in $$^{120}C_{60}$$, is

A
$$1$$

B
$$0$$

C
$$2$$

D
$$3$$

Solution

Exponent of $$5$$ in $$^{ 120 }{ C }_{ 60 }$$
$$={}^{ n }{ C }_{ r }=\dfrac { n! }{ (n-r)!r! } \\ \\ ={}^{ 120 }{ C }_{ 60 }=\dfrac { 120! }{ 60!\times 60! } \\ =\dfrac { 120\times 119\times 118\times .....61 }{ 60\times 59\times 58\times .....2\times 1 } $$
Number of multiples of $$5$$ in Numerator $$=$$ Denominator $$=14$$
Overall exponent of $$5=0$$
Question 4
1 / -0

If $$^nC_{r-1}=36, ^nC_r = 84$$ and $$^nC_{r+1} = 126$$ then the value of $$^nC_8$$ is:

A
$$10$$

B
$$7$$

C
$$9$$

D
$$8$$

Solution

$$^nC_{r - 1} = \cfrac{n!}{(r - 1)! (n - r + 1)!} \ \ ..(1)$$
$$^nC_r = \cfrac{n!}{r!(n - r)!} \ \ ..(2)$$
$$^nC_{r + 1} = \cfrac{n!}{(r + 1)!(n - r - 1)!} \ \ ..(3)$$
Dividing equation $$(2)$$ by $$(1)$$, we get
$$\cfrac{84}{36} = \cfrac{n - r + 1}{r} $$
$$\therefore 7r = 3n - 3r + 3$$
$$\therefore 10r = 3n + 3 \ \ ..(4)$$
Dividing equation $$(3)$$ by $$(2)$$, we get
$$\cfrac{126}{84} = \cfrac{n - r}{r + 1} $$
$$\therefore 3r + 3 = 2n - 2r$$
$$\therefore 2n = 5r + 3 \ \ ..(5)$$
Multiplying equation $$(5)$$ by $$2$$ and adding that to equation $$(4)$$, we get
$$4n + 10r = 3n + 3 + 10r + 6$$
$$\therefore n = 9$$
$$\therefore \displaystyle ^nC_{8} = \ ^{9}C_{8} = 9$$
Question 5
1 / -0

The value of the expression $$^{47}C_4$$ +$$ \sum_{j=1}^{5} { }^{52-j}C_3$$ is

A
$$^{51}C_4$$

B
$$^{52}C_4$$

C
$$^{52}C_3$$

D
$$^{53}C_4$$

Solution

$$^{ 47 }{ C }_{ 4 }+\sum _{ j=1 }^{ 5 }{ ^{ 52-j }{ C }_{ 3 } } =?\\ ^{ 47 }{ C }_{ 4 }+^{ 51 }{ C }_{ 3 }+^{ 50 }{ C }_{ 3 }+^{ 49 }{ C }_{ 3 }+^{ 48 }{ C }_{ 3 }+^{ 47 }{ C }_{ 3 }$$
We Know ,
$$^{ n }{ C }_{ r }+^{ n }{ C }_{ n-r }=^{ n+1 }{ C }_{ r }\\ \Rightarrow ^{ 47 }{ C }_{ 4 }+^{ 47 }{ C }_{ 3 }+^{ 48 }{ C }_{ 3 }+^{ 49 }{ C }_{ 3 }+^{ 50 }{ C }_{ 3 }+^{ 51 }{ C }_{ 3 }\\ \Rightarrow ^{ 48 }{ C }_{ 4 }+^{ 48 }{ C }_{ 3 }+^{ 49 }{ C }_{ 3 }+^{ 50 }{ C }_{ 3 }+^{ 51 }{ C }_{ 3 }\\ \Rightarrow ^{ 49 }{ C }_{ 4 }+^{ 49 }{ C }_{ 3 }+^{ 50 }{ C }_{ 3 }+^{ 51 }{ C }_{ 3 }\\ \Rightarrow ^{ 50 }{ C }_{ 4 }+^{ 50 }{ C }_{ 3 }+^{ 51 }{ C }_{ 3 }\\ \Rightarrow ^{ 51 }{ C }_{ 4 }+^{ 51 }{ C }_{ 3 }\\ \Rightarrow ^{ 52 }{ C }_{ 4 }$$
Hence the correct answer is $$^{ 52 }{ C }_{ 4 }$$
Question 6
1 / -0

If $$^{n-1}C_r = (k^2 - 3) ^nC_{r+1} ,$$ then k belongs to the interval

A
$$[\sqrt{-3}, \sqrt3 ]$$

B
$$(-\infty,-2)$$

C
$$(2, \infty)$$

D
$$(\sqrt3, 2]$$

Solution

$$^{ n-1 }{ C }_{ r }=({ K }^{ 2 }-3)^{ n }{ C }_{ r+1 }\\ \Rightarrow ^{ n }{ C }_{ r }=\cfrac { n! }{ (n-r)!r! } \\ \Rightarrow \cfrac { \cfrac { (n-1)! }{ (n-r-1)!r! } }{ \cfrac { n! }{ (n-r-1)!(r+1)! } } ={ K }^{ 2 }-3\\ \Rightarrow \cfrac { r }{ n } ={ K }^{ 2 }-3$$
We know in the value of $$^{ n }{ C }_{ r }$$
value of $$n\ge r$$
$$\Rightarrow \cfrac { r }{ n } $$ value lies between $$[0,1]$$
$$\Rightarrow { K }^{ 2 }-3\in \left[ 0,1 \right] \\ \Rightarrow { K }^{ 2 }\in \left[ 3,4 \right] \\ \Rightarrow { K }\in \left[ \sqrt { 3 } ,2 \right] $$
Question 7
1 / -0

If $$^{20}C_r = ^{20}C_{r-10}$$, then the value of $$^{18}C_r$$ is:

A
$$4896$$

B
$$5432$$

C
$$816$$

D
$$1632$$

Solution

Given $$^{ 20 }{ C }_{ r }=^{ 20 }{ C }_{ r-10 }$$
We Know
$$^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }\\ \Rightarrow r+(n-r)=n\\ \Rightarrow r+r-10=20\\ \Rightarrow 2r=30\\ \Rightarrow r=15\\ \Rightarrow ^{ 18 }{ C }_{ r }=^{ 18 }{ C }_{ 15 }\\ \Rightarrow \dfrac { 18! }{ 15!\times 3! } \\ \Rightarrow \dfrac { 18\times 17\times 16 }{ 3\times 2\times 1 } \\ \Rightarrow 48\times 17\\ \Rightarrow 816$$
Hence the correct answer is $$816$$
Question 8
1 / -0

The exponent of $$7$$ in $$^{120}C_{50}$$, is

A
$$0$$

B
$$2$$

C
$$4$$

D
None of these

Solution

Exponent of $$7$$ in $$^{ 120 }{ C }_{ 50 }$$

$$^{ n }{ C }_{ r }=\dfrac { n! }{ (n-r)!r! } \\ \\ \Rightarrow ^{ 120 }{ C }_{ 50 }=\dfrac { 120! }{ 70!\times 50! } \\ =\dfrac { 120\times 119\times 118\times .....71 }{ 50\times 49\times 48\times .....2\times 1 } $$

Number of multiples of $$7$$ in Numerator $$=$$ Denominator $$=8$$

Therefore Exponent of $$7$$ is $$0$$
Question 9
1 / -0

For a chess tournament $$13$$ people were selected for quarter finals. Each person plays two matches with the other. How many matches have been held in the whole tournament?

A
$$144$$

B
$$156$$

C
$$185$$

D
$$116$$

Solution

$$13$$ People Each people plays two matches
Selecting two players and letting them play
$$={}^{ 13 }{ P }_{ 2 }$$
$$=\dfrac { 13\times 12\times 11\times 10\times .......3\times 2\times 1 }{ 11! }$$
$$=13\times 12$$
$$=156$$
Hence Number of matches is $$156$$
Question 10
1 / -0

The area of regular polygon of n sides with length of side $$\sqrt{3}$$, where n > 1 and satisfies the relation $$\displaystyle \sum_{r=0}^{n} \frac{n^2-3n+3}{2\ ^nC_r}=\sum_{r=0}^n\frac{r}{^nC_r}$$ is :

A
$$30\sqrt{3}$$

B
$$\dfrac{3\sqrt{3}}{4}$$

C
$$\dfrac{5\sqrt{3}}{4}$$

D
$$\dfrac{3\sqrt{3}}{2}$$

Solution

$$\displaystyle \sum\dfrac r{^nC_r} = \sum\dfrac{n-r}{^nC_r} = \dfrac n2\sum\dfrac1{^nC_r}$$
$$\dfrac{n^2-3n+3}{2} \sum\dfrac1{^nC_r} \Rightarrow n^2-3n+3=n$$
$$n^2-4n+3=0 \Rightarrow n=3 (\because n>1)$$
$$\therefore$$ Area of the equilateral triangle with side $$\sqrt3$$ is $$\dfrac{3\sqrt3}4$$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 45

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Permutations and Combinations Test 45

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SHARING IS CARING

Question 1

$${3}^{4}$$

$${3}^{3}$$

$${4}^{4}$$

$$12$$

$${4}^{3}$$

Solution

Question 2

Less than $$10^{7}$$

Between $$10^{7}$$ and $$10^{8}$$

Between $$10^{8}$$ and $$10^{9}$$

Between $$10^{9}$$ and $$10^{10}$$

More than $$10^{10}$$

Solution

Question 3

$$1$$

$$0$$

$$2$$

$$3$$

Solution

Question 4

$$10$$

$$7$$

$$9$$

$$8$$

Solution

Question 5

$$^{51}C_4$$

$$^{52}C_4$$

$$^{52}C_3$$

$$^{53}C_4$$

Solution

Question 6

$$[\sqrt{-3}, \sqrt3 ]$$

$$(-\infty,-2)$$

$$(2, \infty)$$

$$(\sqrt3, 2]$$

Solution

Question 7

$$4896$$

$$5432$$

$$816$$

$$1632$$

Solution

Question 8

$$0$$

$$2$$

$$4$$

None of these

Solution

Question 9

$$144$$

$$156$$

$$185$$

$$116$$

Solution

Question 10

$$30\sqrt{3}$$

$$\dfrac{3\sqrt{3}}{4}$$

$$\dfrac{5\sqrt{3}}{4}$$

$$\dfrac{3\sqrt{3}}{2}$$

Solution

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