Permutations and Combinations Test 29

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Permutations and Combinations Test 29

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Weekly Quiz Competition

Question 1
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In how many ways a garland can be made from exactly $$10$$ flowers

A
$$10!$$

B
$$9!$$

C
$$2(9!)$$

D
$$\dfrac{9!}{2}$$

Solution

Since $$10$$ things can be permuted along a circle in $$9!$$ ways.
But in a garland anticlockwise and clockwise directions are the same, so we have $$\dfrac{9!}{2}$$.
Hence, option D is correct.
Question 2
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Directions For Questions

In these questions, numbers are placed in the figures on the basis of some rules. One place is vacant which if indicated as'?' Find out the correct alternatives to replace the question mark '?'

...view full instructions

The figure is given.

A
$$144$$

B
$$100$$

C
$$116$$

D
$$169$$

E
$$180$$

Solution

By observation, the no. in the middle circle is the sum of the cube of all the no.s outside the square
$$\implies 1^3+2^3+4^3+6^3=289$$ : Circle 2
$$\implies 1^3+5^3+6^3+7^3=685$$ : Circle 3
Thus, $$1^3+2^3+3^3+4^3=100$$ : Circle 1
Thus, the answer is $$100$$.
Question 3
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In how many way can $$12$$ gentlemen sit around a round table so that three specified gentlemen are always together.

A
$$9!$$

B
$$10!$$

C
$$3! 10!$$

D
$$3! 9!$$

Solution

Let us consider the $$3$$ specific gentlemen as a single person that occupies three times the space.

So, now we have $$10$$ persons who can be made to sit in $$9!$$ ways and in the space for $$3$$ gentlemen we can permute them in $$3!$$ ways.

So, total ways are $$3!\, 9!$$.
Question 4
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An alphabet contains a $$A^{'s}$$ and b $$ B^{'s}$$ . (In all a+b letters ). The number of words each containing all the $$ A^{'s}$$ and any number of $$ B^{'s}$$, is

A
$$^{a+b}C_b$$

B
$$^{a+b+1}C_a$$

C
$$^{a+b+1}C_b$$

D
none of these

Solution
Question 5
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The maximum number of intersection points of n circles and n straight lines , among themselves is 80.The value of n is

A
$$7$$

B
$$6$$

C
$$5$$

D
$$8$$

Solution
Question 6
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If $$\alpha ,\beta , \gamma $$ are three consecutive integers. If these integers are raised to first, second and third positive powers respectively, and added then they form a perfect square, the square root of which is equal to the sum of these integers. Also, $$\alpha < \beta < \gamma $$. Then, $$\gamma$$ is equals to:

A
$$3$$

B
$$14$$

C
$$5$$

D
$$11$$

Solution

Let the numbers be $$n-1,n,n+1$$
As per the given information, we have
$$(n-1)+{ n }^{ 2 }+{ (n+1) }^{ 3 }={ (n-1+n+n+1) }^{ 2 }$$
$$\Rightarrow { n }^{ 3 }-5{ n }^{ 2 }+4n=0$$
$$\Rightarrow n=0,1,4$$
For $$n=0,$$ the numbers are: $$-1,0,1$$ this is out as all the numbers should be positive
For $$n=1,$$ the numbers are: $$0,1,2$$ this is out as all the numbers should be positive (‘0’ can’t be taken as positive)
For $$n=4,$$ the numbers are: $$3,4,5$$
We have got largest number which is $$5$$
Hence, the value of $$\gamma$$ is $$5$$.
Question 7
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In the given question, in four out of the five figures of element $$I$$ is related to element $$II$$ in some particular way. Find out the figure in which the element is not related to element $$II$$.The figures are given.

A

B

C

D

E

Solution

The relation between I and II is:

(1)There is a 'shift' of the whole pattern by one spot.

(2)The first $$3$$ shapes shift by one place.

(3)The $$4^{th}$$ shape does not shift but instead changes into some other shape

(4)The $$5^{th}$$ shape shifts by $$2$$ places.

Option D doesn't follow (3) and (4).
Question 8
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Find the missing number in the circle:

A
$$66$$

B
$$72$$

C
$$71$$

D
$$78$$

Solution

$$3\times 2 - 1, 5\times 2 - 1, 9\times 2 - 1, 17\times 2 - 1 + 39 = 72$$.

$$\therefore$$ The solution is $$72$$.
Question 9
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Find the missing number in above figure :

A
$$35$$

B
$$40$$

C
$$49$$

D
$$53$$

Solution

$$23 + 5 = 28, 45 + 5 = 50, 35 + 5 = 40$$.
$$\therefore$$ The solution is $$40$$.
Question 10
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Find the missing number in above figure :

A
$$P$$

B
$$H$$

C
$$M$$

D
$$L$$

Solution

Position of $$N \rightarrow 14$$ (in alphabetical order)
Position of $$T \rightarrow 20$$
Position of $$B \rightarrow 2$$
$$\therefore$$ Position of $$P \rightarrow 16$$.
$$\therefore$$ The solution is $$P$$.