Sequences and Series Test 1

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Sequences and Series Test 1

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Weekly Quiz Competition

Question 1
1 / -0

All possible number are formed using the digits $$1,1,2,2,2,2,3,4,4$$ taken all at a time. The number of such number in which the odd digits occupy even places is:

A
$$175$$

B
$$162$$

C
$$160$$

D
$$180$$

Solution

Number of such number $$=^4C_3\times \dfrac{3!}{2!}\times \dfrac{6!}{2!4!}=180$$
Question 2
1 / -0

The number of four -digit numbers strictly grater than $$4321$$ that can be formed using the digits $$0,1,2,3,4,5$$ (repetition of digits is allowed) is:

A
$$288$$

B
$$306$$

C
$$360$$

D
$$310$$

Solution

(1) The number of four-digit numbers Strating with 5 is equal to $$6^3 = 216$$
(2) Starting with 44 and 55 is equal to $$36\times 2 = 72$$
(3) Starting with 433, 434 and 435 is equal to $$6\times 3 = 18$$
(4) Remaining numbers are 4322, 4323, 4324, 4325 is equal to 4
so total number are
$$716 + 72 + 18 + 4 = 310$$
Question 3
1 / -0

A group of students comprises of $$5$$ boys and $$n$$ girls. If the number of ways, in which a team of $$3$$ students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is $$1750$$, then $$n$$ is equal to

A
$$25$$

B
$$28$$

C
$$27$$

D
$$24$$

Solution

Number of ways of selecting a team with 1 boy and 2 girls$$=^{5}C_{1}. ^{n}C_{2} $$

Number of ways of selecting a team with 2 boy and 1 girls$$=^{5}C_{2} . ^{n}C_{1} $$

Given $$^{5}C_{1}. ^{n}C_{2} + ^{5}C_{2} . ^{n}C_{1} = 1750$$

$$5\cdot \dfrac {n(n-1)}{2!}+10\cdot n=1750$$

$$n^{2} + 3n = 700$$

$$\therefore n = 25$$.
Question 4
1 / -0

The value of $$(^{21}C_1-^{10}C_1)+(^{21}C_2-^{10}C_2)+(^{21}C_3-^{10}C_3)+(^{21}C_4+^{10}C_4)+......+(^{21}C_{10}-^{10}C_{10})$$ is:

A
$$2^{21}-2^{11}$$

B
$$2^{21}-2^{10}$$

C
$$2^{20}-2^9$$

D
$$2^{20}-2^{10}$$

Solution

$$(^{21}C_1-^{10}C_1)+(^{21}C_2-^{10}C_2)+(^{21}C_3-^{10}C_3)+(^{21}C_4+^{10}C_4)+......+(^{21}C_{10}-^{10}C_{10})$$

$$= ^{21}C_1+^{21}C_2+^{21}C_3+ ..... + ^{21}C_{10} - [2^{10}-1]$$

$$=\cfrac 12 [2 ^{21}C_1 + 2 ^{21}C_2+ 2 ^{21}C_3 + ..... + 2 ^{21}C_{10}] - [2^{10}-1]$$

$$=\cfrac 12 [^{21}C_1 + ^{21}C_2+ .... + ^{21}C_{10} + ^{21}C_{11} + ..... + ^{21}C_{19} + ^{21}C_{20}] - [2^{10}-1]$$

$$=\cfrac 12 [2^{21} - 1 -1 ] - [2^{10}-1]$$

$$=2^{20} - 1 - 2^{10} +1$$

$$=2^{20}-2^{10}$$
Question 5
1 / -0

The figure below shows the network connecting cities A, B, C, D, E and F. The arrows indicate permissible direction of travel. What is the number of distinct paths from A to F?

A
9

B
10

C
11

D
8

Solution

(B) The maximum routes from A to F are listed below.

(1) ABDF (2) ACDF (3) ABF (4) ABEF

(5) ACDEF (6) ABCDEF (7) ABDEF (8) ABCDF
Question 6
1 / -0

Value of 0! is always 1.

A
True

B
False

C
Either

D
Neither

Solution

we know $$1!=1$$
Also
$$n!=n\times (n-1)\times (n-2)........3\times 2\times 1\\ n!=n\times (n-1)!\\ 1!=1(1-1)!\\ 1=1(0)!\\ 0!=1$$
$$0! $$ is always $$1$$
Hence its true that $$0!$$ is always $$1$$
Question 7
1 / -0

Each combination corresponds to many permutations.

A
True

B
False

C
Either

D
Neither

Solution

In combination
Each combination can be considered as a set of selection an order
Each selection has a defined order
They can be considered as a permutation
Each cpmbination corresponds to many permutations
Hence the above statement is true.
Question 8
1 / -0

Area of a triangle formed by the points A(5, 2), B(4, 7) and C(7, -4) is _____.

A
2 sq. units

B
$$2\sqrt2$$ sq. units

C
4 sq. units

D
$$4\sqrt2$$ sq. units

Solution

Area of a triangle ABC = $$ \frac { 1 }{ 2 } \left|[ x_{ 1 }\left( y_{ 2 }-y_{ 3 }

\right) +x_{ 2 }\left( y_{ 3 }-y_{ 1 } \right) +x_{ 3 }\left( y_{ 1 }-y_{ 2 }

\right) \right]|$$
Substituting the given coordinates, we have
=$$ \frac { 1 }{ 2 } \left[ 5 \left(7+4) \right) -\left( 4+2\right) +7\left( 2-7) \right) \right]$$
$$ =\frac { 1 }{ 2 } \left( 5 ( 11) +4\times( -6)
+7\times( -5)\right)$$
$$ =\frac { 1 }{ 2 } \left[(55) - ( 24) - ( 35) \right]$$
$$ =\frac { 1 }{ 2 } \times |-4| = 2$$ square units.
Question 9
1 / -0

Factorial of negative numbers is always greater than 1.

A
True

B
False

C
Either

D
Neither

Solution

Factorial : product of an integer with integer less than it.
Factorial can be interpolated using gamma function and gamma function and
gamma function is not defined for negative integer.
Factorial is not defined for negative integer
Question 10
1 / -0

Choose the correct option for the following.
$$n!=n(n-1)(n-2).....3.2.1$$

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

Factorial : the product of an integer and all integer less than that
$$\therefore n!=n\times (n-1)\times (n-2)............3\times 2\times 1$$

$$\therefore $$ The given statement
$$n!=n\times (n-1)\times (n-2)............3\times 2\times 1$$ is True

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Sequences and Series Test 1

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Sequences and Series Test 1

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SHARING IS CARING

Question 1

$$175$$

$$162$$

$$160$$

$$180$$

Solution

Question 2

$$288$$

$$306$$

$$360$$

$$310$$

Solution

Question 3

$$25$$

$$28$$

$$27$$

$$24$$

Solution

Question 4

$$2^{21}-2^{11}$$

$$2^{21}-2^{10}$$

$$2^{20}-2^9$$

$$2^{20}-2^{10}$$

Solution

Question 5

9

10

11

8

Solution

Question 6

True

False

Either

Neither

Solution

Question 7

True

False

Either

Neither

Solution

Question 8

2 sq. units

$$2\sqrt2$$ sq. units

4 sq. units

$$4\sqrt2$$ sq. units

Solution

Question 9

True

False

Either

Neither

Solution

Question 10

True

False

Ambiguous

Data insufficient

Solution

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