Permutations and Combinations Test 14

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Permutations and Combinations Test 14

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Weekly Quiz Competition

Question 1
1 / -0

Let $$ S= \left\{1,2,3,.......n\right\} $$ and $$ A =\left\{(a, b) \left. \right | 1 \geq a, b \geq n\right\} = S \times S\: $$. A subset $$ B$$ of $$A $$ is said to be a good subset if $$ (x, x) \in B$$ for every $$x \in S.$$ Then the number of good subsets of $$A$$ is

A
$$1$$

B
$$2^{n}$$

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

D
$$2^{n^{2}}$$

Solution

The number of element in $$A$$= $$n\times{n}$$
The number of element of A having $$(x,x)$$ for all $$x\in S$$ = $$n$$
Total remaining element = $$(n\times{n})-n$$
Since Subset $$B$$ must have $$(x,x)$$ for all $$x\in S$$ = $$n$$ so it may have any number of remaining element.
Number of ways of selecting elements from remaining element = $${2}^{(n\times{n})-n}$$
Question 2
1 / -0

If $$(1+x)^n=C_0+C_1x+C_2x^2+ ......+C_nx^n,$$ then the value of $$C_0 + 2C_1+ 3 C_2 + ...... (n + 1) C_n$$ will be

A
$$(n + 2)2^{n - 1}$$

B
$$(n + 1)2^{n }$$

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

D
$$(n + 2)2^{n}$$

Solution

$$C_0 + 2 C_1 + 3C_2 + ..... + (n+1)C_n$$
$$= ^nc_0 + 2^nC_1 + 3^nC_2 + ..... + (n + 1) ^nC_n$$
$$(^n C_0 + ^nC_1 + ^nC_2 + ..... + ^nC_n) + (^nC_1 + 2 ^nC_2 + 3^nC_3 + ..... n. ^nC_n)$$
$$= \displaystyle (1 +1)^n + n + n (n - 1) + 3 \frac{n (n -1)(n - 2)}{3} + .... + n$$
$$= 2^n + n \displaystyle \left [ 1 + (n - 1) + \frac{(n - 1) (n - 2)}{2} + ...... + 1 \right ]$$
$$= 2^n + n [1 + 1]^{n - 1} = 2^n + n.2^{n -1}= (n + 2). 2^{n - 1}$$
Question 3
1 / -0

Forty teams play a tournament. Each of them plays with every other team just once. Each game result is a win for one team. If each team has a $$50\%$$ chance of winnings each game, the number of ways such that at the end of the tournament, every team has won a different number of games is:

A
$$\dfrac{1}{780}$$

B
$$40!2^{780}$$

C
$$40! 3^{780}$$

D
None of these
Question 4
1 / -0

The sum of the series $$\displaystyle \sum_{r=0}^{10} {\;}^{20}C_r$$ is

A
$$2^{20}$$

B
$$2^{19}$$

C
$$2^{19}+\displaystyle \frac {1}{2}^{20}C_{10}$$

D
$$2^{19}-\displaystyle \frac {1}{2}^{20}C_{10}$$

Solution

Simplifying the expression we get
$$\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{10}$$ ...(i)
Now
$$(1+x)^{20}=\:^{20}C_{0}+\:^{20}C_{1}x+\:^{20}C_{2}x^{2}+...\:^{20}C_{20}x^{n}$$
Substituting x=1, we get
$$2^{20}=\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{20}$$
Now
$$\:^{20}C_{r}=\:^{20}C_{n-r}$$
Hence
$$2^{20}=2[\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{9}]+\:^{20}C_{10}$$
Now
$$2^{19}-\dfrac{\:^{20}C_{10}}{2}=\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{9}$$
Adding $$\:^{20}C_{10}$$ on both sides we get
$$2^{19}-\dfrac{\:^{20}C_{10}}{2}+\:^{20}C_{10}=\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{9}+\:^{20}C_{10}$$
$$2^{19}+\dfrac{\:^{20}C_{10}}{2}=\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{9}+\:^{20}C_{10}$$
Question 5
1 / -0

There are 8 buses running from Kota to Jaipur and 10 buses running from Jaipur to Delhi. In how many ways a Person can travel from Kota to Delhi via Jaipur by bus?

A
80

B
8

C
10

D
160

Solution

Let $$E_1$$ be the event of travelling from Kota to Jaipur & $$E_2$$ be the event of travelling from Jaipur to Delhi by the person.
$$E_1$$ can happen in 8 ways and $$E_2$$ can happen in 10 ways.
Since both the events $$E_1$$ and $$E_2$$ are to be happened 12 in order, simultaneously,the number of ways $$=8\times10=80$$
Question 6
1 / -0

If $$\displaystyle^{n}P_{r}= 360$$ and $$^ {n}C_{r}=15$$ then find the value of r

A
$$5$$

B
$$4$$

C
$$3$$

D
$$2$$

Solution

$$^nP_r = \dfrac { n! }{ (n-r)! } = 360 $$-- (1)

$$^nC_r = \dfrac { n! }{ (r!)(n-r)! } = 15 $$ -- (2)

Dividing eqn 1 by 2, we get
$$ => r ! = \dfrac {360}{15} = 24 $$
$$ => r ! = 4 \times 2 \times 3 \times 1 = 4 ! $$

So, $$ r = 4 $$
Question 7
1 / -0

The value of $$\displaystyle ^{ 19 }C_{ 18 }+^{ 19 }C_{ 17 }$$

A
$$1200$$

B
$$2000$$

C
$$190$$

D
None of these

Solution

$$\displaystyle ^{ n }C_{ r }=\frac { n! }{ \left( n-r \right) !r! } $$
$$\displaystyle ^{ 19 }C_{ 18 }+^{ 19 }C_{ 17 }=\frac { 19! }{ 18! } +\frac { 19! }{ 17!2! } $$
$$\displaystyle =19+\frac { 19\times 18 }{ 2 } $$
$$\displaystyle =19+171$$
$$\displaystyle =190$$
Question 8
1 / -0

IF$$\displaystyle ^{n}C_{r}=^{n}P_{r}$$ then r can be____

A
0

B
1

C
3

D
Either(1) or (2)

Solution

Given,
$$ ^nC_r=^nP_r$$
$$ => \dfrac { n! }{ (r!)(n-r)! } = \dfrac { n! }{ (n-r)! } $$
$$ => r ! = 1 $$
$$ => r = 0 $$ or $$ 1 $$ as $$ 0 ! = 1; 1 ! = 1 $$
Question 9
1 / -0

Different calenders for the month of February are made so as to serve for all the coming years. The number of such calenders is

A
$$7$$

B
$$2$$

C
$$14$$

D
None of these

Solution

It has to perform two jobs
1- select number of days in February i.e 28 or 29
2-select first day of month in 7 ways
Total way to perform both jobs=2*7=14
Question 10
1 / -0

Find the value of $$\displaystyle ^{10}C_{10}$$

A
$$2$$

B
$$1$$

C
$$4$$

D
$$6$$

Solution

We have $$ ^nC_r = \dfrac { n! }{ (r!)(n-r)! } $$

So, $$ ^{10}C_{10} = \dfrac { 10! }{ (10!) (10-10)! } =\dfrac { 1! }{ 0! } = 1 $$ Since $$ 1 ! = 0 ! = 1 $$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 14

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Permutations and Combinations Test 14

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SHARING IS CARING

Question 1

$$1$$

$$2^{n}$$

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

$$2^{n^{2}}$$

Solution

Question 2

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

$$(n + 1)2^{n }$$

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

$$(n + 2)2^{n}$$

Solution

Question 3

$$\dfrac{1}{780}$$

$$40!2^{780}$$

$$40! 3^{780}$$

None of these

Question 4

$$2^{20}$$

$$2^{19}$$

$$2^{19}+\displaystyle \frac {1}{2}^{20}C_{10}$$

$$2^{19}-\displaystyle \frac {1}{2}^{20}C_{10}$$

Solution

Question 5

80

8

10

160

Solution

Question 6

$$5$$

$$4$$

$$3$$

$$2$$

Solution

Question 7

$$1200$$

$$2000$$

$$190$$

None of these

Solution

Question 8

0

1

3

Either(1) or (2)

Solution

Question 9

$$7$$

$$2$$

$$14$$

None of these

Solution

Question 10

$$2$$

$$1$$

$$4$$

$$6$$

Solution

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