Permutations and Combinations Test 15

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Permutations and Combinations Test 15

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Weekly Quiz Competition

Question 1
1 / -0

The mid points of three sides of a triangle are (0, 1) (0, 2) and (0, 3). Area of this triangle.

A
3

B
0

C
1

D
None of these

Solution
Question 2
1 / -0

How many six letter words be made out of the letters of $ASSIST$ ? In how many words the alphabets $S$ alternates with other letters ?

A
$120,6$

B
$720,\,12$

C
$120,\,12$

D
$720,\,24$

Solution

The word ASSIST has 3 's' and 3 other characters (AIT).
Place the other 3 characters first. This can be done in 3! ways.
Now there are 4 spaces remaining (1234) mark as 1,2,3,4.
3 's' 5 can be placed at the spaces marked as 1,2,3 or at the spaces marked as 2,3,4 (2)was
Total no.of ways =3! .2 $=3\times 2\times 2=12$
Total no.of ways =5!. $=5\times 4\times 3\times 2=120.$
$\therefore$ so, the answer is 120,12.
Question 3
1 / -0

In how many ways can 5 persons A,B,C,D and E sit around a circular table.

A
120

B
240

C
75

D
60

Solution

Given,

$5$ people

$A,B,C,D,E$

ways of sitting around a table

$A,B,D,E,C$

$B,A,D,E,C$

$A,B,C,E,D$

$B,A,C,E,D$

$\therefore ^5P_4=\dfrac{5!}{(5-4)!}=120$ ways
Question 4
1 / -0

If P (n, n) denotes the number of permutations of n different things taken all at a time then P (n, n ) is also identical to:

A
$$

P ( n - 1 , n - 1 )

$$

B
$$

P ( n , n - 1 )

$$

C
$$

\mathrm { r } ! \mathrm { P } ( \mathrm { n } , \mathrm { n } - \mathrm { r } )

$$

D
$$

( n - r ) \cdot P ( n , r )

$$

Solution
Question 5
1 / -0

Observe the given pattern
$4\times 0+1=01$
$4\times1+2=06$
$4\times 2+3=11$
$4\times 3+4=16$
Then the value of $4\times8+9$ is -----

A
$43$

B
$42$

C
$41$

D
$40$

Solution

If we carefully observe the pattern
in first row it shows four multiply with zero addition one equals one
same for other pattern too
so answer of $4\times 8+9 = 41$
Question 6
1 / -0

When we realize a specific implementation of a pancake algorithm, every move when we find the greatest of the sized array and flipping can be modeled through ____________.

A
Combinations

B
Exponential functions

C
Logarithmic functions

D
Permutations

Solution

Unlike a traditional sorting algorithm, which attempts to sort with the fewest comparisons possible, the goal is to sort the sequence in as few reversals as possible.
Here when we flipping the array or stack, we have to take utmost priority to preserve the order of the list so that that sorting doesnt become invalid. Hence we use permutations, we are ensuring that order matters.
Question 7
1 / -0

The value of $\displaystyle E = \frac { (1+17)(1+\frac { 17 }{ 2 } )(1+\frac { 17 }{ 3 } )......(1+\frac { 17 }{ 19 } ) }{ (1+19)(1+\frac { 19 }{ 2 } )(1+\frac { 19 }{ 3 } ).....(1+\frac { 19 }{ 17 } ) }$ is,

A
$1$

B
$^{36}C_{17}$

C
$\dfrac {2}{19}$

D
$^{36}C_{18}$

Solution

$\displaystyle E = \frac { (1+17)(1+\frac { 17 }{ 2 } )(1+\frac { 17 }{3 } )......(1+\frac { 17 }{ 19 } ) }{ (1+19)(1+\frac { 19 }{ 2 } )(1+\frac { 19 }{ 3 } ).....(1+\frac { 19 }{ 17 } ) }$

$\quad =\displaystyle \frac{\displaystyle \frac{18\cdot 19\cdot 20.............36}{1\cdot 2\cdot 3\cdot 4...............19}}{\displaystyle \frac{20\cdot 21\cdot 22.............36}{1\cdot 2\cdot 3\cdot 4...............17}}$

$\ \ \ =\displaystyle \frac{18\cdot 19\cdot 20.............36}{1\cdot 2\cdot 3\cdot 4...............19}\times \frac{1\cdot 2\cdot 3\cdot 4...............17}{20\cdot 21\cdot 22.............36}$

$\quad = \displaystyle \frac{18\cdot 19\cdot 20.............36}{20\cdot 21\cdot 22.............36}\times \frac{1\cdot 2\cdot 3\cdot 4...............17}{1\cdot 2\cdot 3\cdot 4...............19}$

$$\displaystyle \ \ \ =\frac{18\cdot 19}{1}\times \frac{1}{18\cdot 19}=1$$
Question 8
1 / -0

$20$ persons are invited for a party, then the number of ways in which they and the host be seated at a round table is

A
20!

B
21!

C
22!

D
2.20!

Solution

Total no. of persons including host are $(20+1)$
As they have to sit in round table, the required permutation is $(n-1)!$
$=(20+1-1)!$
$=20!$
Question 9
1 / -0

Plot $(3, 0), (5, 0)$ and $(0, 4)$ on cartesian plane. Name the figure formed by joining these points and find its area.

A
Figure formed is a Rectangle and Area $= 52$ sq. unit

B
Figure formed is a triangle and Area $= 4$ sq. unit

C
Figure formed is a square and Area $= 40$ sq. unit

D
None of these

Solution

The figure is a triangle.
We know that the area of the triangle whose vertices are $\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$ and $(x_{3},y_{3})$ is $\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$
Therefore,
Area $= \dfrac{1}{2} [3(0-4)+5(4-0)+0] = 4$ square units.
Question 10
1 / -0

The area of a triangle with vertices $A (3, 0), B (7, 0)$ and $C (8, 4)$ is

A
$14$

B
$28$

C
$8$

D
$6$

Solution

The area of the $\Delta ABC$ with vertices
$A\left( { x }_{ 1 },{ y }_{ 1 } \right)=(3,0), B\left( { x }_{ 2 },{ y }_{ 2 } \right)=(7,0)\ \&\ C\left( { x }_{ 3 },{ y }_{ 3 } \right)=(8,4)$ is
$Ar.\Delta ABC=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right]$
$=\dfrac { 1 }{ 2 } \left\{ 3\left( 0-4 \right) +7\left( 4-0 \right) +8\left( 0-0 \right) \right\}$ sq.units $=8$ sq. units.
Hence, option C.

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Permutations and Combinations Test 15

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Permutations and Combinations Test 15

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Question 1

3

0

1

None of these

Solution

Question 2

120,6120,6120,6

720, 12720,\,12720,12

120, 12120,\,12120,12

720, 24720,\,24720,24

Solution

Question 3

120

240

75

60

Solution

Question 4

$$P ( n - 1 , n - 1 )$$

$$P ( n , n - 1 )$$

$$\mathrm { r } ! \mathrm { P } ( \mathrm { n } , \mathrm { n } - \mathrm { r } )$$

$$( n - r ) \cdot P ( n , r )$$

Solution

Question 5

434343

424242

414141

404040

Solution

Question 6

Combinations

Exponential functions

Logarithmic functions

Permutations

Solution

Question 7

111

36C17 ^{36}C_{17} 36C17​

219\dfrac {2}{19}192​

36C18 ^{36}C_{18} 36C18​

Solution

Question 8

20!

21!

22!

2.20!

Solution

Question 9

Figure formed is a Rectangle and Area =52= 52=52 sq. unit

Figure formed is a triangle and Area =4= 4=4 sq. unit

Figure formed is a square and Area =40= 40=40 sq. unit

None of these

Solution

Question 10

141414

282828

888

666

Solution

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$120,6$

$720,\,12$

$120,\,12$

$720,\,24$

$$

P ( n - 1 , n - 1 )

$$

$$

P ( n , n - 1 )

$$

$$

\mathrm { r } ! \mathrm { P } ( \mathrm { n } , \mathrm { n } - \mathrm { r } )

$$

$$

( n - r ) \cdot P ( n , r )

$$

$43$

$42$

$41$

$40$

$1$

$^{36}C_{17}$

$\dfrac {2}{19}$

$^{36}C_{18}$

Figure formed is a Rectangle and Area $= 52$ sq. unit

Figure formed is a triangle and Area $= 4$ sq. unit

Figure formed is a square and Area $= 40$ sq. unit

$14$

$28$

$8$

$6$