Permutations and Combinations Test 20

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Permutations and Combinations Test 20

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Weekly Quiz Competition

Question 1
1 / -0

If $$^nC_{12} = ^nC_8$$ then is equal to

A
$$12$$

B
$$26$$

C
$$6$$

D
$$20$$

Solution

$$^{ n }{ C }_{ 12 }=^{ n }{ C }_{ 8 }$$
$$\cfrac { n! }{ 12!(n-12)! } =\cfrac { n! }{ 8!(n-8)! } $$
$$\cfrac { 1 }{ 12\times 11\times 10\times 9\times 8!\left( n-12 \right) ! } =\cfrac { 1 }{ 8!(n-8)\times (n-9)(n-10)(n-11) (n-12)! } $$
$$\Longrightarrow \left( n-8 \right) \left( n-9 \right) \left( n-10 \right) \left( n-11 \right) =12\times 11 \times 10 \times 9$$
$$\therefore n-8=12$$
$$n=8+12$$
$$n=20$$
Question 2
1 / -0

If different words are formed with all the letters of the word 'AGAIN' and are arranged alphabetically among themselves as in a dictionary, the word at the 50th place will be

A
NAAGI

B
NAAIG

C
IAAGN

D
IAANG

Solution

We have the word $$AGAIN$$
The words start from $$AAGIN=4!=24$$
The words start from $$GAAIN=\cfrac{4!}{2!}=12$$
The words start from $$IAAGN=\cfrac{4!}{2!}=12$$
Number of total words$$=48$$
$$\therefore 49th$$ word is $$NAAGI$$ and $$50th$$ word is $$NAAIG.$$
Hence, B is the correct option.
Question 3
1 / -0

Out of $$15$$ points in a plane, n points are in the same straight line, $$445$$ triangles can be formed by joining these points. What is the value of n?

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Number of triangles that can be formed is equal to the number of ways to select 3 non-collinear points.
Number of ways to select 3 points from 15 points $$=^{15}C_3$$
Let n points be collinear.
Number of ways to select 3 points out of the n collinear points$$=^{n}C_3$$
So, Number of ways to select 3 non-collinear points = Number of ways to select 3 points using all the points - Number of ways to select 3 points using the collinear points
So, Number of ways to select 3 non-collinear points $$=^{15}C_3-^{n}C_3$$
So, Number of triangles that can be formed $$=^{15}C_3-^{n}C_3$$
$$\therefore 445=^{15}C_3-^{n}C_3$$
$$\therefore 445=455-^{n}C_3$$
$$\therefore ^{n}C_3=10$$
$$\therefore \dfrac{n!}{3!(n-3)!}=10$$
$$\therefore \dfrac{n(n-1)(n-2)}{6}=10$$
$$\therefore n(n-1)(n-2)=60$$
Solving this equation we get $$n=5$$

The answer is option (C)
Question 4
1 / -0

Given that C(n, r) : C ( n, r + 1)=1 : 2 and C (n, r + 1) : C (n, r + 2) = 2 : 3.
Find $$n.$$

A
$$11$$

B
$$12$$

C
$$13$$

D
$$14$$

Solution

Given : $$\cfrac { _{ r }^{ n }{ C } }{ _{ r+1 }^{ n }{ C } } =\cfrac { 1 }{ 2 }$$

$$\Rightarrow 2\cfrac { n! }{ r!(n-r)! } = \cfrac{n!}{(r+1)!(n-r-1)!} $$

$$ \Rightarrow 3r+2=n$$

Also, $$\cfrac { { _{ r+1 }^{ n }{ C } } }{ _{ r+2 }^{ n }{ C } } =\cfrac { 2 }{ 3 }$$

$$\Rightarrow 3\cfrac { n! }{ (r+1)!(n-r-1)! } =2\cfrac { n! }{ (r+2)!(n-r-2)! } $$

$$ \Rightarrow 3r+6=2n-2r-2$$

$$\Rightarrow 5r+8=2n$$

Substituting $$n$$, we get

$$8+5r=2(3r+2)$$

$$8+5r=6r+4$$

$$ \Rightarrow r=4$$

Solving for $$n$$, we get

$$n=3(4)+2=12+2$$

$$n=14$$

Hence, D is correct.
Question 5
1 / -0

The number of six-digit numbers which have sum of their digits as an odd integer, is

A
$$45000$$

B
$$450000$$

C
$$97000$$

D
$$970000$$

Solution

The first six digit no.$$=100000$$
The last six digit no.$$=999999$$
No. of six digit no.$$=999999-100000+1$$
$$ =900000$$
In every two consecutive numbers, one of them will have sum of their digits as even integer.
For example, $$100000$$ have sum of their digits equal to $$1$$
Then $$100001$$ will have sum of their digits equal to $$2$$ and so on...
So, every alternate number have sum of their digits as even number and similarly the other alternate pair of numbers have their sum of digits as odd integer.
Thus, the number of six-digit numbers which have sum of their digits as odd integer and the number of six-digit numbers which have sum of their digits as even integer are equal.
And we know that, sum of digits of the number is either odd or even.
Thus, the number of six-digit number having sum of their digits as odd integer is half of the total number we have
$$\therefore$$ Out of $$900000$$, half of them will have sum as odd.
Therefore $$450000$$ six digit no. will have sum of their digits as an odd integer.
Question 6
1 / -0

The value of $$\left( \dfrac { ^{ 50 }{ C }_{ 0 } }{ 1 } +\dfrac {^{ 50 }{ C }_{ 2 } }{ 3 } +...+\dfrac {^{ 50 }{ C }_{ 50 } }{ 51 } \right) $$ is

A
$$\dfrac { { 2 }^{ 50 } }{ 51 } $$

B
$$\dfrac { { 2 }^{ 50 }-1 }{ 51 } $$

C
$$\dfrac { { 2 }^{ 50 }-1 }{ 50 } $$

D
None of these

Solution

$$\left( \dfrac { _{ }^{ 50 }{ { C }_{ 0 } } }{ 1 } +\dfrac { _{ }^{ 50 }{ { C }_{ 2 } } }{ 3 } +...+\dfrac { _{ }^{ 50 }{ { C }_{ 50 } } }{ 51 } \right) $$
$$=\left( \dfrac { 1 }{ 1 } +\dfrac { 50\times 49 }{ 3\times 2! } +...+\dfrac { 1 }{ 51 } \right) $$
$$=\dfrac { 1 }{ 51 } \left( \dfrac { 51 }{ 1! } +\dfrac { 51\times 50\times 49 }{ 3! } +...+1 \right) $$
$$=\dfrac { 1 }{ 51 } \left( _{ }^{ 51 }{ { C }_{ 1 } }+_{ }^{ 51 }{ { C }_{ 3 } }+...+_{ }^{ 51 }{ { C }_{ 51 } } \right) $$
$$=\dfrac { 1 }{ 51 } { 2 }^{ 51-1 }=\dfrac { { 2 }^{ 50 } }{ 51 } $$
Question 7
1 / -0

If $$^nC_{r-1}=10, ^nC_r=45$$ and $$^nC_{r+1}=120$$, then $$r$$ equals to

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Using $$\dfrac{{}^nC_r}{{}^nC_{r-1}}=\dfrac{n-r+1}{r}$$
$$\Rightarrow \dfrac{{}^nC_r}{{}^nC_{r-1}}=\dfrac{45}{10}$$ and $$\dfrac{{}^nC_{r+1}}{{}^{n}C_{r}}=\dfrac {120}{45}$$
$$\Rightarrow \dfrac{n-r+1}{r}=\dfrac{9}{2}$$
and $$\dfrac{n-r}{r+1}=\dfrac{8}{3}$$
$$\Rightarrow \dfrac{8}{3}(r+1)+1=\dfrac{9}{2}r$$
$$\Rightarrow 16 r+16+6=27 r$$
$$\Rightarrow 11r=22$$
$$\therefore r=2$$
Question 8
1 / -0

If $$\quad { _{ }^{ n }{ C } }_{ 2 }+{ _{ }^{ n }{ C } }_{ 3 }={ _{ }^{ 6 }{ C } }_{ 3 }$$ and $${ _{ }^{ n }{ C } }_{ x }={ _{ }^{ n }{ C } }_{ 3 },x\neq 3$$ then the value of $$x$$ is

A
$$5$$

B
$$4$$

C
$$2$$

D
$$6$$

E
$$1$$

Solution

Given, $${ _{ }^{ n }{ C } }_{ 2 }+{ _{ }^{ n }{ C } }_{ 3 }={ _{ }^{ n }{ C } }_{ 3 }$$
$$\Rightarrow n=5\left[ \because { _{ }^{ n }{ C } }_{ r }+{ _{ }^{ n }{ C } }_{ r-1 }={ _{ }^{ n+1 }{ C } }_{ r } \right] $$
and $$\quad { _{ }^{ n }{ C } }_{ x }={ _{ }^{ n }{ C } }_{ 3 }\Rightarrow { _{ }^{ 5 }{ C } }_{ x }={ _{ }^{ 5 }{ C } }_{ 3 }$$
$$\quad \Rightarrow x=5-3=2$$
Question 9
1 / -0

The number of positive integers satisfying the inequality $${ _{ }^{ n+1 }{ C } }_{ n-2 }-{ _{ }^{ n+1 }{ C } }_{ n-1 }\le 50$$ is

A
$$9$$

B
$$8$$

C
$$7$$

D
$$6$$

Solution

$${ _{ }^{ n+1 }{ C } }_{ n-2 }-{ _{ }^{ n+1 }{ C } }_{ n-1 }\le 50$$
$$\Rightarrow \cfrac { (n-1)! }{ 3!(n-2)! } -\cfrac { (n+1)! }{ 2!(n-2)! } \le 50$$
$$\Rightarrow (n+1)!\left( \cfrac { n-1-3 }{ (n-1)! } \right) \le 300\quad $$
$$\Rightarrow (n+1)n(n-4)\le 300$$
By trail and error we get $$n\leq8$$ to satisfy the above inequality.
Question 10
1 / -0

Find the term independent of $$x$$ in $${ \left( \cfrac { 3 }{ 2 } { x }^{ 2 }-\cfrac { 1 }{ 3x } \right) }^{ 9 }$$.

A
$$\dfrac {6}{15}$$

B
$$\dfrac {7}{18}$$

C
$$\dfrac {7}{8}$$

D
$$\dfrac {4}{9}$$

Solution

We know that the general term in expansion of $$(a+b)^n$$ is given by $$^nC_{r}a^{n-r}b^{r}$$ where r ranges from $$0 $$ to $$ 9$$.
Here $$a= \cfrac{3}{2}x^2, \ b= -\cfrac{1}{3x} \ \And \ n=9$$
For any term to be independent of $$x$$, coefficient of $$x$$ in $$a^{n+1-r}b^{r-1}$$ should be zero.
$$\Rightarrow \ (x^2)^{9-r}(x^{-1})^{r}=x^0$$
$$\Rightarrow 18-2r-r=0$$
$$\Rightarrow r=6$$
Therefore, the term independent of $$x$$ is $$^9C_6 \left (\cfrac{3}{2}\right)^3 \left (-\cfrac{3}{3}\right)^6=\cfrac{7}{18}$$.

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 20

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Permutations and Combinations Test 20

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SHARING IS CARING

Question 1

$$12$$

$$26$$

$$6$$

$$20$$

Solution

Question 2

NAAGI

NAAIG

IAAGN

IAANG

Solution

Question 3

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 4

$$11$$

$$12$$

$$13$$

$$14$$

Solution

Question 5

$$45000$$

$$450000$$

$$97000$$

$$970000$$

Solution

Question 6

$$\dfrac { { 2 }^{ 50 } }{ 51 } $$

$$\dfrac { { 2 }^{ 50 }-1 }{ 51 } $$

$$\dfrac { { 2 }^{ 50 }-1 }{ 50 } $$

None of these

Solution

Question 7

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 8

$$5$$

$$4$$

$$2$$

$$6$$

$$1$$

Solution

Question 9

$$9$$

$$8$$

$$7$$

$$6$$

Solution

Question 10

$$\dfrac {6}{15}$$

$$\dfrac {7}{18}$$

$$\dfrac {7}{8}$$

$$\dfrac {4}{9}$$

Solution

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