Permutations and Combinations Test 49

Result

Permutations and Combinations Test 49

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Weekly Quiz Competition

Question 1
1 / -0

A
$$6$$

B
$$7$$

C
$$8$$

D
$$9$$

E
$$10$$

Solution
Question 2
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
Question 3
1 / -0

Total number of ways of selecting two numbers from the set $${1,2,3,...90}$$ so that their sum is divisible by $$3$$ is

A
$$885$$

B
$$1335$$

C
$$1770$$

D
$$3670$$

Solution
Question 4
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 5
1 / -0

The number of $$n$$ digit numbers which consists of the digits $$1$$ & $$2$$ only if each digits is to be used atleast once, is equal to $$510$$ then $$n$$ is equal to

A
$$7$$

B
$$8$$

C
$$9$$

D
$$10$$
Question 6
1 / -0

Fill in the missing values.
J V R B
Z 18 24 22 14
L 11 17 15 7
? 9 15 13 5
T 15 21 ? 11

A
X and $$29$$

B
P and $$15$$

C
H and $$19$$

D
F and $$12$$

Solution
Question 7
1 / -0

Find the missing number, if same rule is followed in all the three figures.

A
12

B
16

C
14

D
9

Solution

REF.Image.
Given

If we consider down two number of fig(i)

then their product is $$ \Rightarrow 4\times 16 = 64 $$

Now if we take square root then =$$ \sqrt{64}$$

$$ = 8 \to$$upper are we are getting.

similarly from fig (iii)

down number product $$ \Rightarrow 12 \times27 = 324$$

take square roots $$ = \sqrt{324} = 18 \to$$upper number.

So, similarly from fig(ii)

Down product = $$ 18 \times 8 = 144 $$

so square root = $$ \sqrt{144} = 12 \rightarrow $$ option A
Question 8
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 9
1 / -0

Find the missing number in the given figure?

A
$$25$$

B
$$361$$

C
$$196$$

D
$$144$$

Solution
Question 10
1 / -0

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

A
140

B
196

C
280

D
346

Solution

There would be two cases,

(1) When 4 is selected from first five and rest 6 from remaining 8.
This is done in $$^{5}C_{4}\times ^{8}C_{6}$$ ways.

(2) When all 5 are selected from first five and rest five from remaining 8.
This can be done in $$^{5}C_{5}\times ^{8}C_{5}$$ ways.

$$\therefore $$ A total of $$^{5}C_{4}\times ^{8}C_{6}+^{5}C_{5}\times ^{8}C_{5}$$

$$=5\times \cfrac{8!}{6!2!}+\cfrac{8!}{3!5!}$$

$$=\cfrac{5\times 8\times 7}{2}+\cfrac{8\times 7\times 8}{3\times 2}$$

$$=140+56=196\,ways$$