Permutations and Combinations Test 48

Result

Permutations and Combinations Test 48

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

Sum of series : $$\sum_\limits{r=1}^n (r^2+1)( r! )$$ is _______

A
(n+1) !

B
(n+2) ! - 1

C
n(n + 1) !

D
None of these

Solution
Question 2
1 / -0

There are infinite, alike, blue, red, green and yellow balls. Find the number of ways to select $$10$$ balls.

A
$$286$$

B
$$84$$

C
$$715$$

D
None of these.

Solution
Question 3
1 / -0

Let $$(1+x)^n=C_0+C_1x+C_2x^2+.....+C_nX^n$$. (where $$C_r=^nC_r$$)
On the basis of given information, answer the following question.
$$C_1-3(C_3)+5(C_5)-7(C_7)+$$_______ is?

A
$$n.2^{\left(\dfrac{n-1}{2}\right)}$$

B
$$n\left(2^{\dfrac{(n-1)}{2}}\right)\cos \dfrac{n\pi}{4}$$

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

D
$$n.2^{\dfrac{(n-1)}{2}}\cos\left(\dfrac{(n-1)\pi}{4}\right)$$

Solution
Question 4
1 / -0

Maximum number of common chords of a parabola and a circle can be equal to

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

parabola $$ eq^n : x^2 = 4ay ...(1)$$

circle $$ eq^n : x^2 +y^2 + 2fx + 2gy + h =0 ...(2)$$

substituting (1) in(2) $$ [y = x^2 / 4a]$$

$$ x^2 + \dfrac{x^2}{(4a)^2}+2fx + 2gy +h = 0 $$

This is a degree 4 polynomial and has 4 distinct real roots,

so point of interaction b/w circle & parabola $$ \Rightarrow 4 $$

Since common chords can be formed poly any
$$2$$ points out of $$4$$ point of interaction.

$$ \therefore $$ common chords $$ = ^{4}C_{2} = \dfrac{4\times 3 }{2 \times 1} = \boxed{6}$$
(option C)
Question 5
1 / -0

In a certain examination paper, there are $$n$$ question. For $$j = 1, 2....n$$, there are $$2^{n-j}$$ students who answered $$j$$ or more questions wrongly. If the total number of wrong answers is $$4095$$, then the value of $$n$$ is:

A
$$12$$

B
$$11$$

C
$$10$$

D
$$9$$

Solution
Question 6
1 / -0

How many $$10-digit$$ numbers can be formed by using the digits $$1$$ and $$2$$?

A
$$^{10}{P}_{2}$$

B
$$^{10}{C}_{2}$$

C
$${2}^{10}$$

D
$$10\ !$$

Solution

Each place of a ten digit number can be fixed by any of the two digits. So, the number of ways to form a ten digit number is $${2^{10}}$$.
Question 7
1 / -0

The value of $$\displaystyle\sum^{10}_{r=0}(r)$$ $$^{20}C_r$$ is equal to?

A
$$20(2^{18}+{^{19}}C_{10})$$

B
$$10(2^{18}+{^{19}}C_{10})$$

C
$$20(2^{18}+{^{19}}C_{11})$$

D
$$10(2^{18}+{^{19}}C_{11})$$

Solution

$$\begin{matrix} \sum _{ r=0 }^{ 10 }(r){ \, ^{ 20 } }{ C_{ r } } \\ \Rightarrow 0{ +^{ 20 } }{ C_{ 1 } }+{ 2^{ 20 } }{ C_{ 2 } }+{ 3^{ 20 } }{ C_{ 3 } }+............+{ 10^{ 20 } }{ C_{ 10 } } \\ \Rightarrow 20+20\times 19+\dfrac { { 20\times 19\times 18 } }{ 2 } +.........+10\times \dfrac { { 20! } }{ { 10!\times 10! } } \\ \Rightarrow 20\left[ { 1+19+\dfrac { { 19\times 18 } }{ 2 } +\dfrac { { 19\times 18\times 17 } }{ 6 } +...........+\dfrac { { 10\times 19! } }{ { 10!\times 10! } } } \right] \\ \Rightarrow 20\left[ { 1+19+\dfrac { { 19\times 18 } }{ 2 } +\dfrac { { 19\times 18\times 17 } }{ 6 } +.........+\dfrac { { 19! } }{ { 9!\times 10! } } } \right] \\ \, \, \, \, \, \, 20\left[ { ^{ 19 }{ C_{ 0 } }{ +^{ 19 } }{ C_{ 1 } }{ +^{ 19 } }{ C_{ 2 } }{ +^{ 19 } }{ C_{ 3 } }+..........{ +^{ 19 } }{ C_{ 10 } } } \right] \, \, \, \, Ans. \\ \end{matrix}$$
Question 8
1 / -0

What is the value of $$^nC_n$$?

A
zero

B
$$1$$

C
$$n$$

D
$$n!$$

Solution
Question 9
1 / -0

Let $$S$$ be the set of $$6$$ digits numbers $$a_{1},a_{2},a_{3},a_{4},a_{5},a_{6}$$ (all digits distinct) where $$a_{1}>a_{2}>a_{3}<a_{4}>a_{5}<a_{6}$$. Then $$n(S)$$ is equal to

A
$$210$$

B
$$2100$$

C
$$4200$$

D
$$420$$
Question 10
1 / -0

A library has $$'a'$$ copies of one book, $$'b'$$ copies of each of two books, $$'c'$$ copies of each of three books, and single copy each of $$'d'$$. The total number of ways in which these books can be arranged in a row is

A
$$\dfrac{(a+b+c+d)!}{a!b!c!}$$

B
$$\dfrac{(a+2b+3c+d)!}{a!(b!)^{2}(c!)^{3}}$$

C
$$\dfrac{(a+2b+3c+d)!}{a!b!c!}$$

D
none of these

Solution