Permutations and Combinations Test 17

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Permutations and Combinations Test 17

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Question 1
1 / -0

Directions For Questions

Directions (27-31) : Out of nine cells of a square one cell is left blank and in the rest of the cells numbers are written follow some rule. Get the rule and find out the proper option for the blank cell (?)

...view full instructions

91
64
73
84
76
61
25
60
?

A
66

B
68

C
69

D
71

Solution

In column first $$91+84+25$$$$=$$ $$200$$
In column second $$64+76+60$$ $$=$$$$200$$
In column third $$73+61+66$$ $$=$$200
Question 2
1 / -0

Given: $$\dfrac {20!}{18!}=380$$

A
True

B
False

C
Either

D
Neither

Solution

$$\cfrac { 20! }{ 18! } =?\\ 20!=20\times 19\times 18\times ........3\times 2\times 1\\ =20\times 19\times (18!)\\ \cfrac { 20! }{ 18! } =\cfrac { 20\times 19\times (18!) }{ 18! } =20\times 19=380$$
Hence the value matches so the equation is true.
Question 3
1 / -0

In a class of 80 students it is found that 40 students like Tajmahal and 50 students like Charminar and 18 like both. Then the number of students who do not like neither are.

A
18

B
72

C
50

D
8

Solution

Given that
Total students$$=80$$
Students who like Tajmahal $$=40 \quad n(T)$$
Students who like charminar$$=50 \quad n(C)$$
Students who like both $$=18 \quad n(T\cap C)$$
Say $$'x'$$ like neither then,
$$80=n(T)+n(C)-n(T\cap C)+x\\ \Rightarrow 80=40+50-18+x\\ \Rightarrow x=8$$
Question 4
1 / -0

In a combination, the ordering of the selected objects is immaterial whereas in a permutation, the ordering is essential.

A
True

B
False

C
Either

D
Neither

Solution

Combination of few things .
The order of selected things is not specified .
Combination is basically selection hence ordering in immaterial .
Permtation of few things the order of selected things is specified .
Permutation basically means selecting and arranging .
Hence ordering is essential .
Question 5
1 / -0

Out of nine cells of a square, one cell is left blank and in the rest of the cells, numbers are written follow some rule. Get the rule and find out the proper option for the blank cell
2
72
56
?
0
42
12
20
30

A
4

B
6

C
8

D
10

Solution

The pattern of numbers are as followed
$$1^2-1=0$$
$$2^2-1=3$$
$$3^2-3=6$$
$$4^2-4=12$$
$$5^2-5=20$$
$$6^2-6=30$$
$$7^2-7=42$$
$$8^2-8=56$$
$$9^2-9=72$$
Question 6
1 / -0

Identify the progression:
$$A : 4, 7, 10, 13, 16, 19, 22, 25, .....$$
$$B : 4, 7, 9, 10, 13, 14, .........$$

A
Only $$A$$

B
Only $$B$$

C
Both $$A$$ and $$B$$

D
None of these

Solution

Only sequence $$A$$ is progression as its terms follow same pattern
Question 7
1 / -0

Directions For Questions

Directions (27-31) : Out of nine cells of a square one cell is left blank and in the rest of the cells numbers are written follow some rule. Get the rule and find out the proper option for the blank cell (?)

...view full instructions

4
20
25
27
81
9
11
44
?

A
4

B
16

C
30

D
55

Solution

$$20 \div 4 =5 \Rightarrow 5^2=25$$
$$81 \div 27 =3 \Rightarrow 3^2 = 9$$
$$44 \div 11 =4 \Rightarrow 4^2=16$$
Question 8
1 / -0

The numbers 1, 2, ..., 100 are arranged in the squares of an table in the following way: the numbers 1, ... , 10 are in the bottom row in increasing order, numbers 11, ... ,20 are in the next row in increasing order, and so on. One can choose any number and two of its neighbors in two opposite directions (horizontal, vertical, or diagonal). Then either the number is increased by 2 and its neighbors are decreased by 1, or the number is decreased by 2 and its neighbors are increased by 1. After several such operations the table again contains all the numbers 1, 2, ... , 100. Prove that they are in the original order.

A
$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 appear in their original order.

B
$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 appear in their original order.

C
$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

D
$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

Solution

Label the table entry in the i th row and j th column by $$a_{ ij }$$
where the bottom-left corner is in the first row and first column.
Let $$\displaystyle b_{ij}= 10(i-1)+j$$ be the number originally in the i th row and j th column. Observe that $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
is invariant. Indeed, every time entries
$$\displaystyle a_{mn}, a_{pq}, a_{rs}$$ are changed (with m+r= 2p and n+s= 2q), P increases or decreases by $$\displaystyle b_{mn}-2b_{pq}+b_{rs},$$
But this equals $$\displaystyle 10\left ( \left ( m-1 \right )+ \left ( r-1 \right )-2\left ( p-1 \right )+\left ( n+s-2q \right )\right )= 0.$$
In the beginning $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$ at the end, the entires $$a_{ij}$$ equal the $$b_{ij}$$
In some order,we now have $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
By the rearrangement inequality, this is at least $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
with equality only when each $$a_{ij}=b_{ij}$$
The equality does occur since P is invariant. Therefore the $$a_{ij}$$ do indeed equal the $$b_{ij}$$
in the same order, and thus the entries $$1, 2, ... , 100$$ appear in their original order.
Question 9
1 / -0

Unit digit in the number $$(12357)^{655}$$ is

A
$$1$$

B
$$3$$

C
$$7$$

D
$$9$$

Solution

unit digit of $$(12357)^{655}$$

$$=$$ unit digit of $$(7^{655})=7^{4\times 163+3}=7^{4\times 143}+7^3$$

$$=$$ unit digit is $$(1\times 3)=$$ unit digit is 3.
Question 10
1 / -0

A tea party is arranged for $$16$$ people along two sides of a large table with $$8$$ chairs on each side. Four men to sit on one particular side and two on the other side. In how many ways can they be seated?

A
$$\displaystyle\frac{8!\space8!\space10!}{(4!)^2\space6!}$$

B
$$\displaystyle\frac{8!\space8!\space10!}{6!}$$

C
$$\displaystyle\frac{8!\space8!\space10!}{4!!}$$

D
$$\displaystyle\frac{8!\space8!\space10!}{4!\space6!}$$

Solution

There are $$8$$ chairs on each side of the table. Let sides be represented by $$A$$ and $$B$$. Let four persons sit on side $$A$$, then number of ways arranging $$4$$ persons on $$8$$ chairs on side $$A =

\space ^8P_4$$ and then two persons sit on side $$B$$, then the number of ways arranging $$2$$ persons on $$8$$ chairs on side $$B = \space

^8P_2$$ and arranging the remaining $$10$$ persons in remaining $$10$$ chairs in $$10!$$ ways.
Hence the total number of ways in which the persons can be arranged
$$\quad = \space ^8P_4\times\space^8P_2\times\space10!$$
$$\quad = \displaystyle\frac{8!\space8!\space10!}{4!\space6!}$$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 17

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Permutations and Combinations Test 17

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SHARING IS CARING

Question 1

66

68

69

71

Solution

Question 2

True

False

Either

Neither

Solution

Question 3

18

72

50

8

Solution

Question 4

True

False

Either

Neither

Solution

Question 5

4

6

8

10

Solution

Question 6

Only $$A$$

Only $$B$$

Both $$A$$ and $$B$$

None of these

Solution

Question 7

4

16

30

55

Solution

Question 8

$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 appear in their original order.

$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 appear in their original order.

$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

Solution

Question 9

$$1$$

$$3$$

$$7$$

$$9$$

Solution

Question 10

$$\displaystyle\frac{8!\space8!\space10!}{(4!)^2\space6!}$$

$$\displaystyle\frac{8!\space8!\space10!}{6!}$$

$$\displaystyle\frac{8!\space8!\space10!}{4!!}$$

$$\displaystyle\frac{8!\space8!\space10!}{4!\space6!}$$

Solution

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