Permutations and Combinations Test 21

Result

Permutations and Combinations Test 21

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Write down and simplify:
The $$5^{th}$$ term of $${ \left( \cfrac { { x }^{ \frac { 3 }{ 2 } } }{ { a }^{ \frac { 1 }{ 2 } } } -\cfrac { { y }^{ \frac { 5 }{ 2 } } }{ { b }^{ \frac { 3 }{ 2 } } } \right) }^{ 8 }$$.

A

$$\dfrac {60x^6y^{5}}{a^2b^6}$$

B
$$\dfrac {70x^6y^{10}}{a^3b^5}$$

C
$$\dfrac {70x^5y^{5}}{a^2b^6}$$

D
$$\dfrac {70x^6y^{10}}{a^2b^6}$$

Solution

The $$r^{th}$$ term in expansion of $$(p+q)^n$$ is given by $$^nC_{r-1}p^{n+1-r}q^{r-1}$$

$$\therefore$$ in the given expression

$$p=\cfrac{x^\frac{3}{2}}{a^\frac{1}{2}}, \ q=-\cfrac{y^\frac{5}{2}}{b^\frac{3}{2}}, \ r=5 \ \And \ n=8$$

$$\therefore$$ the $$5^{th}$$ term will be $$^8C_4 { \left( \cfrac{x^\frac{3}{2}}{a^\frac{1}{2}}\right)^4 {\left( - \cfrac{y^\frac{5}{2}}{b^\frac{3}{2}} \right)}^4} = \cfrac{70x^6y^{10}}{a^2b^6}$$
Question 2
1 / -0

The number of positive integer satisfying the inequality $$^{n + 1}C_{n} - {}^{n + 1}C_{n - 1} \leq 100$$ is

A
$$9$$

B
$$8$$

C
$$5$$

D
None of these

Solution

$$^{n + 1}C_{n } - {}^{n + 1}C_{n - 1} \leq 100$$
$$\Rightarrow {}^{n + 1}C_{3} - {}^{n + 1}C_{2} \leq 100$$
$$\Rightarrow \dfrac {(n + 1)n(n - 1)}{6} - \dfrac {(n + 1)n}{2} \leq 100$$
$$\Rightarrow (n + 1)n(n - 1) - 3n(n + 1)\leq 600$$
$$\Rightarrow (n + 1) n(n - 4)\leq 600$$
The values of $$n$$ satisfying this inequality are
$$2, 3, 4, 5, 6, 7, 8, 9$$.
Question 3
1 / -0

If PQRS is a convex quadrilateral with 3, 4, 5 and 6 points marked on side PQ, QR, RS and PS respectively. Then, the number of triangles with vertices on different sides is

A
$$220$$

B
$$270$$

C
$$282$$

D
$$342$$

Solution

Total number of triangles $$=$$ Number of triangles with vertices on sides $$(\text{PQ, QR, RS + QR, RS, PS + RS, PS, PQ + PS, PQ, QR})$$
$$= {}^3C_1 \times {}^4C_1 \times {}^5C_1 + {}^4C_1 \times {}^5C_1 \times {}^6C_1 + {}^5C_1 \times {}^6C_1 \times {}^3 C_1 + {}^6C_1 \times {}^3C_1 \times {}^4C_1$$
$$= 60 + 120 + 90 + 72 = 342$$
Question 4
1 / -0

The number of triangles that can be formed by choosing the vertices from a set of $$12$$ points of which $$7$$ points lie on a line is

A
$$185$$

B
$$175$$

C
$$115$$

D
$$105$$

Solution

We can choose any three points out of 12 points and then subtract the no. of ways choosing three points out of 7 points which are in straight line.

$$\therefore$$Total no. of triangles$$= ^{12}C_3-^7C_3=185$$
Question 5
1 / -0

Find the $$4^{th}$$ term of $${ \left( 9x-\cfrac { 1 }{ 3\sqrt { x } } \right) }^{ 18 }$$.

A
$$16500$$

B
$$18564$$

C
$$16540$$

D
$$32600$$

Solution

We know that the $$r^{th}$$ term in expansion of $$(a+b)^n$$ is given by $$^nC_{r-1}a^{n+1-r}b^{r-1}$$
Here $$a=9x, \ n=18 \ \And \ b=-\cfrac{1}{3\sqrt{x}}$$
$$\therefore$$ the fourth term in expansion of $$\left (9x-\cfrac{1}{3\sqrt{x}}\right)^{18}$$ is $$^{18}C_{12} {x}^{6}\left (-\cfrac{1}{3\sqrt{x}}\right)^{12} =18564$$
Question 6
1 / -0

The arithmetic mean of $${ _{ }^{ n }{ C } }_{ 0 },{ _{ }^{ n }{ C } }_{ 1 },....{ _{ }^{ n }{ C } }_{ n }\quad $$, is

A
$$\cfrac { 1 }{ n } $$

B
$$\cfrac { { 2 }^{ n } }{ n } $$

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

D
$$\cfrac { { 2 }^{ n+1 } }{ n } $$

Solution

Clearly,
Required mean $$=\cfrac { { _{ }^{ n }{ C } }_{ 0 }+{ _{ }^{ n }{ C } }_{ 1 }+...+{ _{ }^{ n }{ C } }_{ n } }{ n } =\cfrac { { 2 }^{ n } }{ n } $$
Question 7
1 / -0

If $$^{n}C_{r - 1} = 36$$ and $$^{n}C_{r} = 84$$, then

A
$$13r - n - 3 = 0$$

B
$$10r - 3n - 30 = 0$$

C
$$10r + 3n - 3 = 0$$

D
$$10r - 3n + 3 = 0$$

E
$$10r - 3n - 3 = 0$$

Solution

Given, $$^{n}C_{r - 1} = 36$$
and $$^{n}C_{r} = 84$$
Now, $$\dfrac {^{n}C_{r - 1}}{^{n}C_{r}} = \dfrac {36}{84}$$
$$\Rightarrow \dfrac {r}{n - r + 1} = \dfrac {36}{84} = \dfrac {3}{7}$$
$$\Rightarrow 7r = 3n - 3r + 3$$
$$\Rightarrow 10r = 3n + 3$$
$$\Rightarrow 10r - 3n - 3 = 0$$.
Question 8
1 / -0

If $$\displaystyle \sum _{ k=0 }^{ 18 }{ \cfrac { k }{ { _{ }^{ 18 }{ C } }_{ k } } } =a\sum _{ k=0 }^{ 18 }{ \cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ k } } } $$, then the value of $$a$$ is

A
$$3$$

B
$$9$$

C
$$6$$

D
$$18$$

E
$$36$$

Solution

Given $$\displaystyle \sum _{ k=0 }^{ 18 }{ \cfrac { k }{ { _{ }^{ 18 }{ C } }_{ k } } } =a\sum _{ k=0 }^{ 18 }{ \cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ k } } } $$
$$LHS= \displaystyle\sum _{ k=0 }^{ 18 }{ \cfrac { k }{ { _{ }^{ 18 }{ C } }_{ k } } } =\left[ 0+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 1 } } +\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 2 } } +...+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 18 } } \right] ...(i)\quad $$
$$RHS=a\displaystyle \sum _{ k=0 }^{ 18 }{ \cfrac { k }{ { _{ }^{ 18 }{ C } }_{ k } } } =a\left[ 1+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 1 } } +\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 2 } } +...+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 18 } } \right] ...(ii)\quad $$
From Eq(i) and (ii)
$$18\left( 1+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 1 } } +\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 2 } } +...+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 18 } } \right) +\cfrac { 9 }{ { _{ }^{ 18 }{ C } }_{ 9 } } $$
$$=2a\left( 1+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 1 } } +\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 2 } } +...+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 18 } } \right) +\cfrac { a }{ { _{ }^{ 18 }{ C } }_{ 9 } } $$
$$\therefore$$ $$2a=18$$
$$\Rightarrow$$ $$a=9$$
Question 9
1 / -0

There are $$10$$ persons including $$3$$ ladies. A committee of 4 persons including atleast one lady is to be formed. The number of ways of forming such a committee is :

A
$$160$$

B
$$170$$

C
$$180$$

D
$$175$$

E
$$155$$

Solution

A committee pf $$4$$ persons can be formed in following ways:
(i) $$3$$ persons and one ladies
$$ =^7C_3 \times ^3C_1 = 35 \times 3 = 105 $$
(ii) $$2$$ persons and two ladies
$$ = ^7C_2 \times ^3C_2 = 21 \times 3 = 63 $$
(iii) $$1$$ persons and three ladies
$$ = ^7C_1 \times ^3C_3 = 7 \times 1 = 7 $$
Therefore, total number of ways
$$ = 105 + 63 + 7 = 175 $$
Question 10
1 / -0

The number of ways of distributing $$8$$ identical balls in $$3$$ distinct boxes, so that none of the boxes is empty is

A
$$5$$

B
$$21$$

C
$${ 3 }^{ 8 }$$

D
$$^{ 8 }{ { C }_{ 3 } }$$

Solution

The total number of ways of dividing $$8$$ identical balls in $$3$$ distinct boxes, so that one of the boxes is empty, is $$^{ 8-1 }{ { C }_{ 3-1 } }$$.
That is further equal to $$^{ 7 }{ { C }_{ 2 } }=\dfrac { 7! }{ 2!5! } =21$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 21

Result

Permutations and Combinations Test 21

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\dfrac {60x^6y^{5}}{a^2b^6}$$

$$\dfrac {70x^6y^{10}}{a^3b^5}$$

$$\dfrac {70x^5y^{5}}{a^2b^6}$$

$$\dfrac {70x^6y^{10}}{a^2b^6}$$

Solution

Question 2

$$9$$

$$8$$

$$5$$

None of these

Solution

Question 3

$$220$$

$$270$$

$$282$$

$$342$$

Solution

Question 4

$$185$$

$$175$$

$$115$$

$$105$$

Solution

Question 5

$$16500$$

$$18564$$

$$16540$$

$$32600$$

Solution

Question 6

$$\cfrac { 1 }{ n } $$

$$\cfrac { { 2 }^{ n } }{ n } $$

$$\cfrac { { 2 }^{ n-1 } }{ n } $$

$$\cfrac { { 2 }^{ n+1 } }{ n } $$

Solution

Question 7

$$13r - n - 3 = 0$$

$$10r - 3n - 30 = 0$$

$$10r + 3n - 3 = 0$$

$$10r - 3n + 3 = 0$$

$$10r - 3n - 3 = 0$$

Solution

Question 8

$$3$$

$$9$$

$$6$$

$$18$$

$$36$$

Solution

Question 9

$$160$$

$$170$$

$$180$$

$$175$$

$$155$$

Solution

Question 10

$$5$$

$$21$$

$${ 3 }^{ 8 }$$

$$^{ 8 }{ { C }_{ 3 } }$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.