Permutations and Combinations Test 46

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Permutations and Combinations Test 46

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Weekly Quiz Competition

Question 1
1 / -0

How many ways $$ 5 $$ persons can sit around the round table :

A
$$ 120 $$

B
$$ 24 $$

C
$$ 60 $$

D
$$ 12 $$

Solution

Total number of ways to sitting $$ 5 $$ persons around the round table.
$$ =(5-1)! = 4! $$
$$ = 4 \times 3 \times 2 \times 1 = 24 $$
hence option (B) is correct.
Question 2
1 / -0

The number of distinct rational numbers x such that $$\displaystyle 0 < x < 1$$ and $$\displaystyle x = \frac{p}{q}r$$, where $$\displaystyle p,q \epsilon \left \{ 1,2,3,4,5,6 \right \}$$,is

A
$$15$$

B
$$13$$

C
$$12$$

D
$$11$$

Solution

Since total number of digits $$=6$$
So to get a number of a number of the form $$\dfrac { p }{ q } $$ we will have to choose two numbers out of the 6 numbers.
There is only one way to arrange them since the number $$x$$ has to satisfy 0 Hence no. $$x = \left( \begin{matrix} 6 \\ 2 \end{matrix} \right) \times 1=15$$, but some numbers have to be removed because they are repeated.
They are :
$$\dfrac { 2 }{ 4 } =\dfrac { 1 }{ 2 } \\ \dfrac { 4 }{ 6 } =\dfrac { 2 }{ 3 } \\ \dfrac { 2 }{ 6 } =\dfrac { 1 }{ 3 } \\ \dfrac { 3 }{ 6 } =\dfrac { 1 }{ 2 } $$
Thus the number of removed numbers$$=4$$
Therefore, total no. of distinct $$x = 15-4=11$$
Hence, option 'D' is correct.
Question 3
1 / -0

The number of rectangles that can be obtained by joining four of the twelve vertices of a $$12$$ sided regular polygon is

A
$$66$$

B
$$30$$

C
$$24$$

D
$$15$$

Solution

The first vertex can be choosed in $$12$$ ways and diagonally opposite to it is $$1$$ vertex. Now for $$3rd$$ vertex we have $$10$$ choices and for $$4th$$ $$1.$$
However, each rectangle is counted $$8$$ times.
$$\therefore$$ No. of ways $$=\dfrac{12\times1\times10\times1}{8}$$ $$=15$$ ways.
Hence, the answer is $$15.$$
Question 4
1 / -0

Three vertices are chosen randomly from the seven vertices of a regular $$7$$ -sided polygon. The probability that they form the vertices of an isosceles triangle is

A
$$\displaystyle \frac{1}{7}$$

B
$$\displaystyle \frac{1}{3}$$

C
$$\displaystyle \frac{3}{7}$$

D
$$\displaystyle \frac{3}{5}$$

Solution

A regular 3 sided polygon is nothing else but a heptagon for creating isosceles triangle we need to choose adjacent sides only.
No. of $$\triangle 's$$ formed $$=7{ { C }_{ 3 } }$$
While number of isosceles triangle formed $$=$$ No. of points $$\times $$ points available $$=7\times 3.$$
$$\Rightarrow$$ So, probability $$=\dfrac { 7\times 3 }{ 7{ { C }_{ 3 } } } = \dfrac { 21 }{ 35 } =\dfrac { 3 }{ 5 } .$$
Hence, the answer is $$\dfrac { 3 }{ 5 }.$$
Question 5
1 / -0

The number of ways in which three numbers in A.P. can be seleced from the set of first n natural number if n is odd is

A
$$ \displaystyle \frac{n\left ( n-2 \right )}{4} $$

B
$$ \displaystyle \frac{n\left ( n-1 \right )^2}{4} $$

C
$$ \displaystyle \frac{\left ( n-1 \right )^2}{4} $$

D
None of these

Solution

In order to solve this question, we must observe the number of ways in which we can select the first term of the required A.P. for different values of the common difference($$r$$) starting from $$r=1$$ given that there are only $$3$$ terms before $$n$$.
For $$r=1$$, the number of ways in which we can select the first term of the A.P. $$=n-2$$
For $$r=2$$, the number of ways to select the first term $$=n-4$$
For $$r=3$$, the number of ways to select the first term $$=n-6$$
Now we see a pattern emerging, we also realize from this that $$r<=\dfrac{n-1}{2}$$ for an A.P. with 3 terms to exist in the given interval.
$$\therefore$$ The final answer $$=n-2+n-4+n-6+...+5+3+1$$
Now we use the formula to find the sum of an A.P. which is $$S_n=\dfrac{n}{2}[a_1+a_n]$$
$$\therefore$$ Answer $$=\dfrac{n-1}{4}[1+n-2]=\dfrac{(n-1)^2}{4}$$
Question 6
1 / -0

The sides of a quadrilateral are all positive integers and three of them are $$5, 10, 20.$$ How many possible value are there for the fourth side?

A
$$29$$

B
$$31$$

C
$$32$$

D
$$34$$

Solution
Question 7
1 / -0

A cabinet of ministers consists of 11 ministers, one minister being the chief minister. A meeting is to be held in a room having a round table and 11 chairsround it, one of them being meant for the chairman. The number of ways in which the ministers can take their chairs, the chief minister occupying the chairman's place, is

A
$$\displaystyle \frac{1}{2} (10!)$$

B
$$\displaystyle 9!$$

C
$$\displaystyle 10!$$

D
none of these

Solution

These are circular permutations
so,
$$=(n-1)!$$ = $$10!$$
Question 8
1 / -0

In a test there were n questions. In the test $$\displaystyle 2^{n-i}$$ students gave wrong answers to i questions where $$\displaystyle i=1,2,3...,n$$. If the total number of wrong answers given is 2047 then n is

A
12

B
11

C
10

D
none of these

Solution

Total number of wrong answers $$= 2^{n-1}+2^{n-2}+...+2^{2}+2+1$$
$$=2^{n-1}+2^{n-2}+...+2^{4}+2^{3}+2^{2}+2+1$$
$$\Rightarrow 2^{n}-1=2043$$ [Using formula for sum of G.P]
$$ \Rightarrow 2^{n}=2048$$
$$ \Rightarrow 2^{n}=2^{11}$$
$$\Rightarrow n = 11$$
Hence, option 'B' is correct.
Question 9
1 / -0

If the difference of the number of arrangements of three things from a certain number of dissimilar things and the number of selections of the same number of things from them exceeds $$ 100$$, then the least number of dissimilar things is

A
$$8$$

B
$$6$$

C
$$5$$

D
$$7$$

Solution

Here,
$$^nP_3-^nC_3 > 100$$

or $$\dfrac {n!}{(n-3)!}-\dfrac {n!}{3!(n-3)!} > 100$$

or $$\dfrac {5}{6}n(n-1)(n-2) > 100$$

or $$n(n-1)(n-2) > 100$$

or $$n(n-1)(n-2) > 6\times 5\times 4$$

or $$n=7, 8, .....$$
Question 10
1 / -0

If four points are $$A(6,3),B(-3,5),C(4,-2)$$ and $$P(x,y),$$ then the ratio of the areas of $$\triangle PBC$$ and $$\triangle ABC$$ is:

A
$$\displaystyle \frac { x+y-2 }{ 7 } $$

B
$$\displaystyle \frac { x-y-2 }{ 7 } $$

C
$$\displaystyle \frac { x-y+2 }{ 2 } $$

D
$$\displaystyle \frac { x+y+2 }{ 2 } $$

Solution

Given: Coordinates of points $$\displaystyle A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 6,3 \right) ,B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -3,5 \right) ,C\left( { x }_{ 3, }{ y }_{ 3 } \right) =\left( 4,-2 \right) $$ and $$\displaystyle P\left( x,y \right) .$$

We know that the area of:
$$\displaystyle\triangle PBC=\frac { 1 }{ 2 } \left[ x\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 3 }\left( y-{ y }_{ 2 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-y \right) \right] $$

$$\displaystyle=\frac { 1 }{ 2 } \left[ x\left( 5+2 \right) +4\left( y-5 \right) -3\left( -2-y \right) \right] $$ $$\displaystyle=\frac { 1 }{ 2 } \left[ 7x+7y-14 \right] $$

Similarly, the area of

$$\displaystyle\triangle ABC=\frac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) \right] +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) $$

$$\displaystyle=\dfrac{1}{2}[6\left( 5+2 \right) -3\left( -2-3 \right) +4\left( 3-5 \right)] =\dfrac{49}{2}$$

Therefore, the ratio of the areas of $$\triangle PAB$$ and $$\triangle ABC$$

$$\displaystyle=\frac { 7x+7y-14 }{ 49 } =\frac { 7\left( x+y-2 \right) }{ 49 } =\frac { x+y-2 }{ 7 } $$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 46

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Permutations and Combinations Test 46

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SHARING IS CARING

Question 1

$$ 120 $$

$$ 24 $$

$$ 60 $$

$$ 12 $$

Solution

Question 2

$$15$$

$$13$$

$$12$$

$$11$$

Solution

Question 3

$$66$$

$$30$$

$$24$$

$$15$$

Solution

Question 4

$$\displaystyle \frac{1}{7}$$

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{3}{7}$$

$$\displaystyle \frac{3}{5}$$

Solution

Question 5

$$ \displaystyle \frac{n\left ( n-2 \right )}{4} $$

$$ \displaystyle \frac{n\left ( n-1 \right )^2}{4} $$

$$ \displaystyle \frac{\left ( n-1 \right )^2}{4} $$

None of these

Solution

Question 6

$$29$$

$$31$$

$$32$$

$$34$$

Solution

Question 7

$$\displaystyle \frac{1}{2} (10!)$$

$$\displaystyle 9!$$

$$\displaystyle 10!$$

none of these

Solution

Question 8

12

11

10

none of these

Solution

Question 9

$$8$$

$$6$$

$$5$$

$$7$$

Solution

Question 10

$$\displaystyle \frac { x+y-2 }{ 7 } $$

$$\displaystyle \frac { x-y-2 }{ 7 } $$

$$\displaystyle \frac { x-y+2 }{ 2 } $$

$$\displaystyle \frac { x+y+2 }{ 2 } $$

Solution

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