Permutations and Combinations Test 35

Result

Permutations and Combinations Test 35

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

Set of value of r for which, $$^{18}C_{r-2}+2\cdot {^{18}C_{r-1}}+{^{18}C_{r}} \geq {^{20}C_{13}}$$ contains?

A
$$4$$ elements

B
$$5$$ elements

C
$$7$$ elements

D
$$10$$ elements

Solution
Question 2
1 / -0

The no.of triangles formed by selecting the points from Regular pentagon is

A
$$10$$

B
$$12$$

C
$$16$$

D
$$none$$

Solution

The no .of points in pentagon is $$5$$
The points required to form rectangle is $$3$$
The no .of ways of selecting is $$^5C_3=10$$
Question 3
1 / -0

$$^{n}C_{r}+2^{n}C_{r+1}+^{n}C_{r+2}$$ is equal to

A
$$2.^{n}C_{r+2}$$

B
$$^{n+1}C_{r+1}$$

C
$$^{n+2}C_{r+2}$$

D
$$none\ of\ these$$

Solution
Question 4
1 / -0

Value of $$\displaystyle \sum _{ r=0 }^{ n} r.\left(^{n}C_{r}\right)^{2}$$ is equal to

A
$$n.{^{2n}C_{r}}$$

B
$$\dfrac{n.{^{2n}C_{r}}}{2}$$

C
$$n^{2}.{^{2n}C_{r}}$$

D
$$\dfrac{n^{2}.{^{2n}C_{r}}}{2}$$
Question 5
1 / -0

The no .of ways of selecting a prime numbers from First 10 natural numbers is

A
$$^{10}C_4$$

B
$$^4C_{10}$$

C
$$^{10}P_4$$

D
$$^{10}C_5$$

Solution

The prime numbers are $$2,3,5,7$$
No.of numbers $$=4$$
Total no .of numbers $$10$$
No .of ways to select is given as $$^{10}C_4$$
'
Question 6
1 / -0

A rectangle with sides 2m - 1 and 2n - 1 is divided into square of unit length by drawing parallel lines as shown in diagram, then the number of rectangles possible with odd side length is

A
$$(m+n-1)^2$$

B
$$4^{m+n-1}$$

C
$$m^2-n^2$$

D
$$m(m+1)n(n+1)$$

Solution

REF.Image
A rectangle with sides 2m-1 and 2n-1 is divide into square units length by drawing parallel lines as shown in diagram, then the number of rectangle possible wide odd side length
Sol:
There are 2m vertical (number 1,2........2m) and 2n horizontal lines (numbered 1,2.......2n).
To form the required rectangle we must selected two horizontal lines, one even numbered and one odd numbered and similarly two vertical lines.
The number of rectangle is then
$$(1+3+5+......+(2m-1))(1+3+5+......+(2n-1))-m^{2}.n^{2}$$
$$\therefore $$ so, the answer is $$c.m^{2}-n^{2}$$
Question 7
1 / -0

How many integers are there such that $$2 \le n \le 100$$ and the highest common factor of $$n$$ and $$36$$ is $$1$$?

A
$$166$$

B
$$332$$

C
$$331$$

D
$$416$$

Solution

$$36.2^{2}.3^{2}$$
Since HCF (36,n) =1
Therefore, n sholud not be a multiple of 2 or 3.
$$ 2 \leq n\leq 100$$
Total numbers = T= 1000.
Number of numbers divisible by 2= $$N_{2}$$
Number of numbers divisible by 3= $$N_{3}$$
Number of numbers divisible by 6= $$N_{6}$$
Therefore, there are $$T-N_{2}-N_{3}+N_{6}$$ total integers
a=2
l=1000
d=2
We know, $$l=a+(N_{2}-1)d.$$
$$N_{2}=\frac{1000-2}{2}+1$$
$$N_{2}=\frac{998}{2}+1$$
$$N_{2}=499+1=500$$
Similarly,
$$N_{3}=\frac{999-3}{3}+1=333$$
$$N_{6}=\frac{999-6}{3}+1=166$$
Answer
= 999-500-333+166
= 332
Question 8
1 / -0

If $$^mC_3+^mC_4>^{m+1}C_3$$, then least value of $$m$$ is :

A
6

B
7

C
5

D
None of these

Solution

$$^mC_3+^mC_4>^{m+1}C_3$$
Expanding the given
$$\frac{m!}{(m-3)!\cdot 3!}+\frac{m!}{(m-4)!\cdot 4!}>\frac{(m+1)!}{(m-2)!\cdot 3!}$$
Dividing m! and multiplying $$(m-4)!\cdot 3!$$
$$\frac{1}{(m-3)}+\frac{1}{4}>\frac{m+1}{(m-2)(m-3)}$$
Further solving
$$4(m-2)+(m-2)(m-3)>4(m+1)$$
$$m^2-5m-6>0$$
$$(m-6)(m+1)>0$$
$$\therefore m>6,m>-1$$
m cannot be -ve so, m>6
so answer will be 7
Answer
Question 9
1 / -0

The value of $$\left( \begin{matrix} 30 \\ 0 \end{matrix} \right) \left( \begin{matrix} 30 \\ 10 \end{matrix} \right) -\left( \begin{matrix} 30 \\ 1 \end{matrix} \right) \left( \begin{matrix} 30 \\ 11 \end{matrix} \right) +\left( \begin{matrix} 30 \\ 2 \end{matrix} \right) \left( \begin{matrix} 30 \\ 12 \end{matrix} \right) .....+\left( \begin{matrix} 30 \\ 20 \end{matrix} \right) \left( \begin{matrix} 30 \\ 30 \end{matrix} \right) $$ is, where $$\left( \begin{matrix} n \\ r \end{matrix} \right) =^{ n }{ C }_{ r }.$$

A
$$\left( \begin{matrix} 30 \\ 10 \end{matrix} \right) $$

B
$$\left( \begin{matrix} 30 \\ 15\end{matrix} \right) $$

C
$$\left( \begin{matrix} 60 \\ 30\end{matrix} \right) $$

D
$$\left( \begin{matrix} 31\\ 10 \end{matrix} \right) $$

Solution
Question 10
1 / -0

The number of ways in which $$9$$ persons can be divided into three equal groups, is

A
$$1680$$

B
$$840$$

C
$$560$$

D
$$280$$

Solution