Permutations and Combinations Test 9

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Permutations and Combinations Test 9

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Question 1
1 / -0

We number both the rows and the columns of an $$8$$ $$\times$$ $$8$$ chess-board with the numbers $$1$$ to $$8$$. A number of grains are placed onto each square, in such a way that the number of grains on a certain square equals the product of its row and column numbers. How many grains are there on the entire chessboard?

A
$$1296$$

B
$$1096$$

C
$$2490$$

D
$$1156$$

Solution

In the first column, we have, successively,
$$1 \times 1, 1 \times 2, 1 \times 3, ..... , 1 \times 8$$ grains.
So, in the first column : $$ 1 \times (1+2+3+4+5+6+7+8)$$
In the second column : $$ 2 \times (1+2+3+4+5+6+7+8)$$
In the third column : $$ 3 \times (1+2+3+4+5+6+7+8).$$
Finally, in the eighth column : $$ 8 \times (1 + 2 + 3+ 4+5+6+7+8)$$
Since $$1+2+3+4+5+6+7+8 = (1/2)\times 8 \times(1+8) = 36$$, there are $$36^2 =1296$$ grains on the board.
Question 2
1 / -0

A hall has 12 gates. In how many ways can a man enter the hall through one gate and come out through a different gate?

A
$$144$$

B
$$132$$

C
$$121$$

D
$$156$$

Solution

The person can enter the hall in 12 gates and exit from any of the other 11 gates.
Hence the total number of ways $$ = 12 \times 11 =132 $$
Question 3
1 / -0

$$\displaystyle \frac{^{\alpha}C_{2}}{^{\alpha + 1}C_{4}}$$ =

A
$$\displaystyle\frac{1}{3}$$

B
$$\displaystyle\frac{12}{(\alpha+1)(\alpha-2)}$$

C
$$\displaystyle\frac{4}{(\alpha+1)(\alpha-3)}$$

D
$$\displaystyle\frac{12}{(\alpha+1)(\alpha-2)(\alpha-3)}$$

Solution

$$=\dfrac{^{\alpha}C_2}{^{\alpha + 1}C_4}$$

$$=\dfrac{\alpha !4!(\alpha -3)!}{(\alpha -2)!2!(\alpha +1)!}$$

$$=\dfrac{\alpha !4.3.2!(\alpha -3)!}{(\alpha -2)(\alpha -3)!2!(\alpha +1)\alpha !}$$

$$=\dfrac{12}{(\alpha -2)(\alpha +1)}$$

Hence, option 'B' is correct.
Question 4
1 / -0

The total number of permutation of $$n$$ different things taken not more than $$r$$ at a time, where each thing may be repeated any number of times, is

A
$$\displaystyle \frac { n\left( { n }^{ n }-1 \right) }{ n-1 } $$

B
$$\displaystyle \frac { \left( { n }^{ r }-1 \right) }{ n-1 } $$

C
$$\displaystyle \frac { n\left( { n }^{ r }-1 \right) }{ n-1 } $$

D
$$\displaystyle \frac { \left( { n }^{ n }-1 \right) }{ n-1 } $$

Solution

One place can be filled in $$n$$ ways by $$n$$ things, the first and second places can be filled in $$=n\times n={ n }^{ 2 }$$ ways.
The first, second and third places can be filled in $$=n\times n\times n={ n }^{ 3 }$$
Similarly, the first, second,....rth places can be filled in $$=n\times n\times n\times ......\times n\left( r\ times \right) ={ n }^{ r }$$
So required total number $$=n+{ n }^{ 2 }+{ n }^{ 3 }+.....+{ n }^{ r }=\displaystyle\frac { n\left( { n }^{ r }-1 \right) }{ n-1 } $$ (Sum of a GP)
Question 5
1 / -0

The number of different ways in which five "alike dashes" and eight "alike dots" can be arranged using only seven of these "dashes" and "dots" is

A
$$350$$

B
$$120$$

C
$$1287$$

D
None of these

Solution

Dashes
Dots
Arrangements
$$5$$
$$2$$
$$^7C_2$$
$$4$$
$$3$$
$$^7C_3$$
$$3$$
$$4$$
$$^7C_4$$
$$2$$
$$5$$
$$^7C_5$$
$$1$$
$$6$$
$$^7C_6$$
$$0$$
$$7$$
$$^7C_7$$
The total number of ways is $$^7C_2+^7C_3+^7C_4+^7C_5+^7C_6=2^7-8=120$$.
Question 6
1 / -0

There are three stations $$A,\space B$$ and $$C$$, five routes for going from station $$A$$ to station $$B$$ and four routes for going from station $$B$$ to station $$C$$.
Find the number of different ways through which a person can go from station $$A$$ to $$C$$ via $$B$$

A
10

B
15

C
20

D
25

Solution

Since there are five routes for going from station A to station B
So number of ways of going from A to B is $$5$$
And there are four routes for going from B to C, so the number of ways of going from B to C is $$4$$
Thus using multiplication rule, number of ways of going from A to C via B is, $$5\times 4=20$$
Question 7
1 / -0

Given six line segments of length $$2, 3, 4, 5, 6, 7$$ units. Then the number of triangles that can be formed by joining these lines is

A
$$^6C_3 - 7$$

B
$$^6C_3 - 1$$

C
$$^6C_3$$

D
$$^6C_3 - 2$$

Solution

Given lengths of line segments are $$2,3,4,5,6,7$$,
The number ways of selecting $$3$$ lengths from them is $$_{ }^{ 6 }{ { C }_{ 3 } }$$,
Consider the cases in which a triangle is not formed,
i.e., when any one of these two conditions is not satisfied,
$$1)$$sum of any two lengths is greater than the third length
$$2)$$difference between any two lengths is smaller than the third one,
Now,
number of such cases are
$$2,3,5$$
$$2,3,6$$
$$2,3,7$$
$$2,4,6$$
$$2,4,7$$
$$2,5,7$$
$$3,4,7$$
$$\therefore$$ the total number of ways are $$_{ }^{ 6 }{ { C }_{ 3 } }-7$$
Question 8
1 / -0

Number of ways 6 rings can be worn on four fingers of one hand?

A
$$4095$$

B
$$4096$$

C
$$4097$$

D
$$4098$$

Solution

These are rings so, repetition is allowed
Number of ways for first ring are $$4$$, for two rings $$4\times 4$$...
So for $$6$$ rings there are $$4^{6} ways = 4096$$ ways
Hence, option 'B' is correct.
Question 9
1 / -0

The value of $$\displaystyle ^{ 47 }{ { C }_{ 4 } }+\sum _{ r=1 }^{ 5 }{ ^{ 52-r }{ { C }_{ 3 } } } $$ is equal to

A
$$^{ 47 }{ { C }_{ 6 } }$$

B
$$^{ 52 }{ { C }_{ 5 } }$$

C
$$^{ 52 }{ { C }_{ 4 } }$$

D
none of these

Solution

We need value of $$={ _{ }^{ 47 }{ C } }_{ 4 }^{ }+\sum _{ r=1 }^{ 5 }{ { _{ }^{ 52-r }{ C } }_{ 3 }^{ } } $$
$$={ _{ }^{ 47 }{ C } }_{ 4 }^{ }+{ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C } }_{ 3 }^{ }+{ _{ }^{ 49 }{ C } }_{ 3 }^{ }+{ _{ }^{ 48 }{ C } }_{ 3 }^{ }+{ _{ }^{ 47 }{ C } }_{ 3 }^{ }$$
$$={ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C } }_{ 3 }^{ }+{ _{ }^{ 49 }{ C } }_{ 3 }^{ }+{ _{ }^{ 48 }{ C } }_{ 3 }^{ }+{ _{ }^{ 47 }{ C } }_{ 3 }^{ }+{ _{ }^{ 47 }{ C } }_{ 4 }={ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C } }_{ 3 }^{ }+{ _{ }^{ 49 }{ C } }_{ 3 }^{ }+{ _{ }^{ 48 }{ C } }_{ 3 }^{ }+{ _{ }^{ 48 }{ C } }_{ 4 }^{ }$$
$$={ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C} }_{ 3 }^{ }+{ _{ }^{ 49 }{ C } }_{ 3 }^{ }+{ _{ }^{ 49 }{ C } }_{ 4 }^{ }={ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C } }_{ 4 }^{ }$$
$$={ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 51 }{ C } }_{ 4 }^{ }={ _{ }^{ 52 }{ C } }_{ 4 }^{ }$$
Question 10
1 / -0

If $$m$$ parallel lines in a plane are intersected by a family of $$n$$ parallel lines, the number of parallelograms than can be formed is

A
$$\dfrac {1}{4} mn(m - 1)(n - 1)$$

B
$$\dfrac {1}{2} mn (m - 1)(n - 1)$$

C
$$\dfrac {1}{4} m^{2}n^{2}$$

D
None of these

Solution

The number of selection of two parallel from $$m$$ lines is $$^{m}C_{2}$$.
The number of selection of two parallel lines from $$n$$ lines is $$^{n}C_{2}$$.
Hence, the number of parallelograms lines is
$$^{m}C_{2} \times ^{n}C_{2} = \dfrac {1}{4} mn (m - 1)(n - 1)$$.

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Re-Attempt Weekly Quiz Competition

Dashes	Dots	Arrangements
$$5$$	$$2$$	$$^7C_2$$
$$4$$	$$3$$	$$^7C_3$$
$$3$$	$$4$$	$$^7C_4$$
$$2$$	$$5$$	$$^7C_5$$
$$1$$	$$6$$	$$^7C_6$$
$$0$$	$$7$$	$$^7C_7$$

Permutations and Combinations Test 9

Result

Permutations and Combinations Test 9

Score

-

Rank

-

SHARING IS CARING

Question 1

$$1296$$

$$1096$$

$$2490$$

$$1156$$

Solution

Question 2

$$144$$

$$132$$

$$121$$

$$156$$

Solution

Question 3

$$\displaystyle\frac{1}{3}$$

$$\displaystyle\frac{12}{(\alpha+1)(\alpha-2)}$$

$$\displaystyle\frac{4}{(\alpha+1)(\alpha-3)}$$

$$\displaystyle\frac{12}{(\alpha+1)(\alpha-2)(\alpha-3)}$$

Solution

Question 4

$$\displaystyle \frac { n\left( { n }^{ n }-1 \right) }{ n-1 } $$

$$\displaystyle \frac { \left( { n }^{ r }-1 \right) }{ n-1 } $$

$$\displaystyle \frac { n\left( { n }^{ r }-1 \right) }{ n-1 } $$

$$\displaystyle \frac { \left( { n }^{ n }-1 \right) }{ n-1 } $$

Solution

Question 5

$$350$$

$$120$$

$$1287$$

None of these

Solution

Question 6

10

15

20

25

Solution

Question 7

$$^6C_3 - 7$$

$$^6C_3 - 1$$

$$^6C_3$$

$$^6C_3 - 2$$

Solution

Question 8

$$4095$$

$$4096$$

$$4097$$

$$4098$$

Solution

Question 9

$$^{ 47 }{ { C }_{ 6 } }$$

$$^{ 52 }{ { C }_{ 5 } }$$

$$^{ 52 }{ { C }_{ 4 } }$$

none of these

Solution

Question 10

$$\dfrac {1}{4} mn(m - 1)(n - 1)$$

$$\dfrac {1}{2} mn (m - 1)(n - 1)$$

$$\dfrac {1}{4} m^{2}n^{2}$$

None of these

Solution

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