Permutations and Combinations Test 27

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Permutations and Combinations Test 27

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Weekly Quiz Competition

Question 1
1 / -0

In the $$xy$$-plane, the vertices of a triangle are $$(-1,3), (6,3)$$ and $$(-1,-4)$$. The area of the triangle is ___ square units.

A
$$10$$

B
$$17.5$$

C
$$24.5$$

D
$$35$$

Solution

If $$(x_1,y_1)$$,$$(x_2,y_2)$$ and $$(x_3,y_3)$$ are the vertices of a triangle then its area is given by,
$$A=\left| \dfrac { 1 }{ 2 } (x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 })) \right| $$
Therefore, with the vertices $$(-1,3)$$, $$(6,3)$$ and $$(-1,-4)$$.
Area of triangle is given by,
$$A=\left| \dfrac { 1 }{ 2 } (-1(3-(-4))+6(-4-3)+(-1)(3-3)) \right|$$
$$ =\left| \dfrac { 1 }{ 2 } (-7-42) \right| $$
$$=\left| \dfrac { 1 }{ 2 } (-49) \right| $$
$$=\left| -24.5 \right|$$
$$ =24.5$$ square units.
Question 2
1 / -0

In how many ways $$7$$ men and $$7$$ women can be seated around a round table such that no two women can sit together?

A
$$(7!)^2$$

B
$$7!\times 6!$$

C
$$(6!)^2$$

D
$$7!$$

Solution

Lets first place the men (M). * here indicates the linker of round table *M -M - M - M - M* which is in $$(7-1)!$$ ways = $$6!$$
So we have to place the women in between the men which is on the $$5$$ empty seats So $$7$$ women can sit on $$7$$ seats in $$(7)!$$ ways
So the answer is $$7!\times 6!$$
Question 3
1 / -0

Let $$\boxed { n }$$ be defined as $$\frac{(n+2)!}{(n-1)!}$$, what is the value of $$\frac{\boxed{7}}{\boxed {3}}$$ ?

A
4.4

B
8.4

C
12.4

D
16.4

E
20.4

Solution

Given, box $$n$$ is equal to $$\dfrac{(n+2)!}{(n-1)!}=(n+2)(n+1)n$$
We get box $$7$$ is equal to $$9 \times 8 \times 7 =504$$
We get box $$3$$ is equal to $$5 \times 4 \times 3 = 60$$
Therefore, If we divide those two, we get $$\dfrac{504}{60} = 8.4$$.
Question 4
1 / -0

Which of the following is correct?

A
When the positions are numbered, then the circular permutations are treated as a linear arrangement.

B
In linear arrangements, it does not make difference whether the positions are numbered are not.

C
Both A and B

D
None of these.

Solution

By numbering position we are actually fixing staring point which makes it similar to linear permutation.so option A is correct
Linear permutation already has a staring point so it dose not need seats to be numbered so option B is correct
Question 5
1 / -0

Two vertices of a triangle are (2, 1) and (3, -2). Its third vertex is (x, y) such that $$\displaystyle y=x+3$$. If its area is 5 sq. units, what are the co-ordinates of the third vertex?

A
(3.5, -6.5)

B
(3.5, 6.5)

C
(-1.5, -1.5)

D
(1.5, -1.5)

Solution

Given third vertex such that,
$$y=(x+3)$$
since, Area of triangle $$ABC=55q$$ units
$$\pm \dfrac{1}{2}{x(1+2)+2(-2-y)+3(y-1)}=\xi$$
$$\Rightarrow \pm \dfrac{1}{2}{x+2x-4-2y+3y-3}=\xi$$
$$\Rightarrow {3x+y-7}=\pm10$$
$$\Rightarrow 3x+y-17=0$$ ------------(1)
and $$3x+y+3=0$$ ------------(2)
Given that $$A(x,y)$$ lies any $$=(x+3)$$ ------------(3)
from equation (L) and (3),
$$x=\dfrac{7}{2},y=\dfrac{13}{2}$$
$$\Rightarrow \boxed{x=3.\xi\,and \, y=6.\xi}$$
from equation (2) and (3) we get,
$$x=\dfrac{-3}{2}\,and\, y=1.\xi $$
Question 6
1 / -0

Twelve friends go out for a dinner to $$UTSAV$$ restaurant. There are two circular tables one with $$7$$ chairs and one with $$5$$ chairs. In how many ways can the group settle down themselves for dinner?

A
$$\dfrac{12!}{7!\times5!}$$

B
$$\dfrac{12!}{35}$$

C
$$12!$$

D
$$12! 5! 7!$$

Solution

Number of ways of selecting $$7$$ people for first round table $$={}^{12}C_5$$
Number of ways of selecting remaining $$5$$ people for second round table $$={}^{5}C_5$$
Arranging $$7$$ people on first round table $$=(7-1)!=6!$$
Arranging $$5$$ people on second round table $$=(5-1)!=4!$$
Hence the answer is $$={}^{12}C_5\times {}^{5}C_5\times 6! \times 4!$$
$$=\dfrac{{12}!\times 5! \times 6! \times 4!}{7! \times 5!\times 5!\times 0!}=\dfrac{12!}{7 \times 5}=\dfrac{12!}{35}$$
Hence the correct answer is $$\dfrac{12!}{35}$$.
Question 7
1 / -0

If $$n$$ books can be arranged on an ordinary shelf in $$720$$ ways, then in how many ways an these books be arranged in a circular shelf ?

A
$$120$$

B
$$720$$

C
$$360$$

D
$$60$$

Solution

$$n$$ books can be arranged in $$N!$$ ways
Given: $$n!=720=6!$$
Therefore, $$n=6$$
The number of ways books can be arranged in circular shelf $$=(n-1)!=5! =5\times 4\times 3\times 2\times 1=120$$
Hence the correct answer is $$120$$.
Question 8
1 / -0

Area of the triangle formed by the points $$\left( 0,0 \right) ,\left( 2,0 \right) $$ and $$\left( 0,2 \right) $$ is

A
$$1$$ sq. unit

B
$$2$$ sq. units

C
$$4$$ sq. units

D
$$8$$ sq. units

Solution

Given: $$A=(x_1,y_1)=(0,0)$$

$$B=(x_2,y_2)=(2,0)$$ and

  $$C=(x_3,y_3)=(0,2)$$

Area of triangle $$=\dfrac {1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)++x_3(y_1-y_2)]$$

$$=\dfrac {1}{2}[0(0-2)+2(2-0)+0(0-0)]$$

         $$=\dfrac {1}{2}[2(2)]$$

         $$=2$$ sq. units.
Question 9
1 / -0

Find the value of $$a$$ if area of the triangle is $$17$$ square units whose vertices are $$(0,0), (4,a), (6,4)$$.

A
$$a=-2$$

B
$$a=-3$$

C
$$a=3$$

D
none of the above

Solution

vertices of the triangle are $$A(0,0), B(4,a), C(6,4)$$.

Then area of triangle$$ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$

$$\Rightarrow 17=\dfrac{1}{2}[0(a-4)+4(4-0)+6(0-a)$$

$$\Rightarrow 34=16-6a$$

$$\Rightarrow a=-\dfrac{18}{6}=-3$$
Question 10
1 / -0

Out of $$7$$ consonants and $$4$$ vowels, words are formed each having $$3$$ consonants and $$2$$ vowels. The number of such words that can be formed is

A
$$210$$

B
$$25200$$

C
$$2520$$

D
$$302400$$

Solution

Total no. of consonants is $$7$$ and of vowels is $$4$$
Out of $$7$$ consonants, $$3$$ are chosen to form a word. So, this can be done in $${ }^{7}C_{3}$$ ways
Out of $$4$$ vowels, $$2$$ are chosen to form a word. So, that can be done in $${ }^{4}C_{2}$$ ways
So, the no. of select $$3$$ consonants and $$2$$ vowels is $$^{7}C_{3} \times ^{4}C_{2}$$
Now, we can arrange the word containing $$5$$ letters in $$5!$$ ways.
Thus, the total no. of words formed each having $$3$$ consonants and $$2$$ vowels is
$$^{7}C_{3} \times ^{4}C_{2}\times 5! = 25200$$
Hence, the answer is $$25200$$.

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Permutations and Combinations Test 27

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Permutations and Combinations Test 27

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SHARING IS CARING

Question 1

$$10$$

$$17.5$$

$$24.5$$

$$35$$

Solution

Question 2

$$(7!)^2$$

$$7!\times 6!$$

$$(6!)^2$$

$$7!$$

Solution

Question 3

4.4

8.4

12.4

16.4

20.4

Solution

Question 4

When the positions are numbered, then the circular permutations are treated as a linear arrangement.

In linear arrangements, it does not make difference whether the positions are numbered are not.

Both A and B

None of these.

Solution

Question 5

(3.5, -6.5)

(3.5, 6.5)

(-1.5, -1.5)

(1.5, -1.5)

Solution

Question 6

$$\dfrac{12!}{7!\times5!}$$

$$\dfrac{12!}{35}$$

$$12!$$

$$12! 5! 7!$$

Solution

Question 7

$$120$$

$$720$$

$$360$$

$$60$$

Solution

Question 8

$$1$$ sq. unit

$$2$$ sq. units

$$4$$ sq. units

$$8$$ sq. units

Solution

Question 9

$$a=-2$$

$$a=-3$$

$$a=3$$

none of the above

Solution

Question 10

$$210$$

$$25200$$

$$2520$$

$$302400$$

Solution

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