Permutations and Combinations Test 36

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Permutations and Combinations Test 36

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=1-x+x^2-x^3+....-x^{15}+x^{16}+x^{17}$$, then the coefficient of $$x^2$$ in $$f(x-1)$$ is?

A
$$826$$

B
$$816$$

C
$$822$$

D
None of these

Solution

$$f\left( x \right) = 1 - \left( {x - 1} \right) + {\left( {x - 1} \right)^2} - {\left( {x - 1} \right)^3} + {\left( {x - 1} \right)^4}.... - {\left( {x - 1} \right)^{15}} + {\left( {x - 1} \right)^{16}} + {\left( {x - 1} \right)^{17}}$$
The coefficient of $${x^2}$$ by using binomial expansion
$$^2{c_0}{x^2}{ + ^3}{c_1}{x^2}{ + ^4}{c_2}{x^2}{ + ^5}{c_3}{x^2}{ + ^6}{c_4}{x^2}{ + ^7}{c_5}{x^2}{ + ^8}{c_6}{x^2}{ + ^9}{c_7}{x^2}$$
$${ + ^{10}}{c_8}{x^2}{ - ^{11}}{c_9}{x^2}{ + ^{12}}{c_{10}}{x^2}{ - ^{13}}{c_{11}}{x^2}{ + ^{14}}{c_{12}}{x^2}{ - ^{15}}{c_{13}}{x^2}{ + ^{16}}{c_{14}}{x^2}{ - ^{17}}{c_{15}}{x^2}$$
$$ \Rightarrow {x^2}\left( {1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 + 91 + 105 + 120 + 136} \right)$$
= 372 + 444 = 816
Question 2
1 / -0

Replace the question mark (?) with the correct option.

A
$$2$$

B
$$7$$

C
$$4$$

D
$$9$$

Solution
Question 3
1 / -0

Let $$5 < n_1 < n_2 < n_3 < n_4$$ be integers such that $$n_1+n_2+n_3+n_4=35$$. The number of such distinct arrangements $$(n_1, n_2, n_3, n_4)$$.

A
$$^{38}C_3$$

B
$$^8C_3$$

C
$$5$$

D
$$6$$

Solution

$${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+.....+{ n }_{ k }=R$$
for this arrangement,
No. of arrangement or distinct arrangements are $${ R+K-1 }_{ { C }_{ K-1 } }$$
So, for $${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+{ n }_{ 4 }=35$$
No. of arrangements $$={ 35+4-1 }_{ { C }_{ 4-1 } }={ 38 }_{ { C }_{ 3 } }$$
Question 4
1 / -0

If $$p:q:r=1:2:3\;then$$ $$\sqrt {5{p^2} + {q^2} + {r^2}} $$ is equal to:

A
4r

B
5p

C
2q

D
5

Solution

Consider the problem

Given, $$p:q:r=1:2:3$$

Therefore, $$\frac{p}{1} = \frac{q}{2} = \frac{r}{4} = k$$ (let)

So, $$p = k,q = 2k,r = 4k$$

Therefore,

$$\begin{array}{l} \sqrt { 5{ p^{ 2 } }+{ q^{ 2 } }+{ r^{ 2 } } } \\ =\sqrt { 5{ k^{ 2 } }+4{ k^{ 2 } }+16{ k^{ 2 } } } \\ =\sqrt { 25{ k^{ 2 } } } \\ 5k=5p \end{array}$$

Hence, Option $$B$$ is the correct answer which is $$5p$$
Question 5
1 / -0

Area of the triangle formed by $$(x_{1,}y_{1}),(x_{2},y_{2}), (3y_{2},(-2y_{1}))$$

A
$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

B
$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{3} \right] $$

C
$$ \dfrac{2}{3}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

D
$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{3}} \right] $$

Solution

We have,

$$ A\left( {{x}_{1}},{{y}_{1}} \right)=\left( {{x}_{1}},{{y}_{1}} \right) $$

$$ B\left( {{x}_{2}},{{y}_{2}} \right)=\left( {{x}_{2}},{{y}_{2}} \right) $$

$$ C\left( {{x}_{3}},{{y}_{3}} \right)=\left( 3{{y}_{2}},-2{{y}_{1}} \right) $$

We know that the area of triangle is

$$ Area\,of\,\Delta ABC=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}+2{{y}_{1}} \right)+{{x}_{2}}\left( -2{{y}_{1}}-{{y}_{1}} \right)+3{{y}_{2}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$
Hence, the is the answer
Question 6
1 / -0

$$7$$ boys and $$8$$ girls have to sit in a row on $$15$$ chairs numbered from $$1$$ to $$15$$ then?

A
Number of ways boys and girls sit alternately is $$8!7!$$

B
Number of ways boys and girls sit alternately is $$2(8!7!)$$

C
The number of ways in which first and fifteenth chair are occupied by boys and between any two boys an even number of girls sit is $$^9C_4$$ $$8!7!$$

D
The number of ways in which first and last seat are occupied by boys and between any two boys an even number of girls sit is $$(2{^9C_4}8!7!)$$.

Solution

$$\begin{matrix} G-denotes\, \, Girls \\ X-denotes\, \, boys \\ \Rightarrow G\times G\times G\times G\times G\times G\times G\times G \\ \Rightarrow Number\, \, of\, \, ways\, \, of\, Girls\, \, is\, \, 8!\, \, alternate \\ \Rightarrow Number\, \, of\, \, ways\, \, of\, boys\, \, is\, \, 7! \\ \, \, \, \, \, \, \, after\, \, late \\ \Rightarrow Number\, \, of\, \, boys\, \, and\, \, girls\, \, sit\, \, alternate\, \, =8!7! \\ When\, \, first\, \, and\, \, fifteen\, \, chair\, \, are\, \, fixed\, \, and\, \\ \, first\, \, chair\, \, \times G\times G\times G\times G\times G\times G\times G\times fifteen\, \, chair\, \, between\, \, any\, \, two\, \, boys \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \underline { 9{ C_{ 4 } }8!7! } \, \, \, Ans. \\ \end{matrix}$$
Question 7
1 / -0

A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the element of P.A subset Q of A is again chosen. The number of way of choosing P and Q so that P Q =$$\phi $$ is :-

A
$$2^{2n}-^{2n}C_{n}$$

B
$$2^{n}$$

C
$$2^{n}-1$$

D
$$3^{n}$$

Solution

Let A = $${a_{1},a_{2},a_{3},....a_{n}}$$. For $$a_{1}$$ $$\epsilon $$ For $$a_{1} \epsilon $$A we have following choices:
(i) $$a_{1} \epsilon $$ P and $$a_{1} \epsilon $$ Q
(ii) $$a_{1} \epsilon $$P $$\notin $$
(iii) $$_{1}\notin P and a_{1}\epsilon Q$$
(Iv)$$_{1}\notin P and a_{1}\notin Q$$
Out of these only (Ii), and (iii) and (iv) imply $$a_{1}\notin P\cap Q $$ therefore, the number of ways in which none of $$a_{1},a_{2}.....a_{n}$$ belong $$P\cap Q$$ is $$3^{n}$$.
Question 8
1 / -0

A box contains 5 pairs of shoes. If 4 shoes are selected, then the number of ways in which exactly one pair of shoes obtained is :

A
120

B
140

C
160

D
180

Solution

We have,

A box contains pair of shoes $$=5$$

Selected pair of shoes $$=4$$

Then,

Exactly one pair of shoes obtained is

$$ {{=}^{5}}{{P}_{4}} $$

$$ =\dfrac{5!}{\left( 5-4 \right)!} $$

$$ =\dfrac{5!}{1!}=5! $$

$$ =5\times 4\times 3\times 2\times 1 $$

$$ =120 $$
Hence, this is the answer.
Question 9
1 / -0

There are n locks and n matching keys. If all the locks and keys are to be perfectly matched, find the maximum number of trails required to open a lock.

A
$$^{n}C_{2}$$

B
$$\sum _{ k=2 }^{ n }{ \left( k+2 \right) } $$

C
$$\dfrac{n(n+1)}{2}$$

D
$$^{n+1}C_{2}$$

Solution
Question 10
1 / -0

Find the area of the triangle formed by the mid points of sides of the triangle whose vertices are $$(2, 1)$$, $$(-2, 3)$$, $$(4, -3)$$.

A
$$1.5$$ sq. units

B
$$3$$ sq. units

C
$$6$$ sq. units

D
$$20$$ sq. units

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 36

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Permutations and Combinations Test 36

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Question 1

$$826$$

$$816$$

$$822$$

None of these

Solution

Question 2

$$2$$

$$7$$

$$4$$

$$9$$

Solution

Question 3

$$^{38}C_3$$

$$^8C_3$$

$$5$$

$$6$$

Solution

Question 4

4r

5p

2q

5

Solution

Question 5

$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{3} \right] $$

$$ \dfrac{2}{3}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{3}} \right] $$

Solution

Question 6

Number of ways boys and girls sit alternately is $$8!7!$$

Number of ways boys and girls sit alternately is $$2(8!7!)$$

The number of ways in which first and fifteenth chair are occupied by boys and between any two boys an even number of girls sit is $$^9C_4$$ $$8!7!$$

The number of ways in which first and last seat are occupied by boys and between any two boys an even number of girls sit is $$(2{^9C_4}8!7!)$$.

Solution

Question 7

$$2^{2n}-^{2n}C_{n}$$

$$2^{n}$$

$$2^{n}-1$$

$$3^{n}$$

Solution

Question 8

120

140

160

180

Solution

Question 9

$$^{n}C_{2}$$

$$\sum _{ k=2 }^{ n }{ \left( k+2 \right) } $$

$$\dfrac{n(n+1)}{2}$$

$$^{n+1}C_{2}$$

Solution

Question 10

$$1.5$$ sq. units

$$3$$ sq. units

$$6$$ sq. units

$$20$$ sq. units

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