Permutations and Combinations Test 47

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Permutations and Combinations Test 47

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Weekly Quiz Competition

Question 1
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Directions For Questions

The area of a triangle whose vertices are $$\displaystyle \left( { x }_{ 1 },{ y }_{ 1 } \right) ,\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ and $$\displaystyle \left( { x }_{ 3 },{ y }_{ 3 } \right) $$ is given by $$\displaystyle \Delta =\frac { 1 }{ 2 } \left| { x }_{ 1 }\left( { y }_{ 2 },{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 },{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 },{ y }_{ 2 } \right) \right| $$. The points $$\displaystyle \left( { x }_{ 1 },{ y }_{ 1 } \right) ,\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ and $$\displaystyle \left( { x }_{ 3 },{ y }_{ 3 } \right) $$ are collinear of $$\displaystyle \Delta =0$$

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Determine the area of the triangle whose vertices are $$\displaystyle \left( \frac { 1 }{ 2 } ,\frac { -1 }{ 2 } \right) ,\left( 2,\frac { -1 }{ 2 } \right) $$ and $$\displaystyle \left( 2,\frac { \sqrt { 3 } -1 }{ 2 } \right) $$.

A
$$\displaystyle 3\sqrt { 3 } $$

B
$$\displaystyle \frac { 3\sqrt { 3 } }{ 8 } $$

C
$$\displaystyle 12\sqrt { 3 } $$

D
$$\displaystyle \frac { 3\sqrt { 3 } }{ 4 } $$

Solution
Question 2
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In how many ways can $$12$$ people be seated around a table?

A
$$11!$$

B
$$12!$$

C
$$\dfrac{12!}{2}$$

D
$$1$$

Solution

Number of people $$=12$$
When in circular arrangement
The possible number of arrangements of $$n$$ objects around a table (circular) $$(n-1)!$$
Therefore number of ways of arranging $$12$$ people
$$\Rightarrow (12-1)!$$
$$\Rightarrow 11!$$ Ways
Therefore number of ways $$=11!$$
Question 3
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How many necklaces of $$10$$ beads each can be made from $$20$$ beads of different colors?

A
$$19!$$

B
$$\dfrac{19!}{10!}$$

C
$$\dfrac{19!}{10!^2}$$

D
$$20!$$

Solution

$$10$$ Beads $$\longrightarrow $$ Each necklece
$$20$$ Beads total
Number of ways of arranging $$20$$ Beads in circular Form $$=19!$$
There should be $$2$$ sets of necklace of $$10$$ beads each
Number of ways $$=10!\times 10!$$
Total ways
$$=\cfrac { 19! }{ 10!\times 10! } \\ =\cfrac { 19! }{ { (10!) }^{ 2 } } $$
Question 4
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Numbers can be classified into two categories,depending on their divisible conditions.
They are (i) Even numbers $$(2p) \vee p \epsilon N$$ (ii) odd numbers $$(2p + 1) \vee p \epsilon N$$
a. $$a_1, a_2 ...... a_{2013}$$ are integers, not necessarily distinct.
$$x = (-1)^{a_1}+(-1)^{a_2}+.....+(-1)^{a_{1006}}$$
$$y = (-1)^{a_{1007}}+(-1)^{a_{1008}}+......+(-1)^{a_{2013}}$$

Then which of the following is true?

A
$$(-1)^x = 1; (-1)^y = 1$$

B
$$(-1)^x = 1; (-1)^y = -1$$

C
$$(-1)^x = -1; (-1)^y = 1$$

D
$$(-1)^x = -1; (-1)^y = -1$$

Solution
Question 5
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If $$ { _{ }^{ n }{ C } }_{ 4 },{ _{ }^{ n }{ C } }_{ 5 }$$ and $$ { _{ }^{ n }{ C } }_{ 6 }$$ are in AP, then $$n$$ is

A
$$7$$ or $$14$$

B
$$7$$

C
$$14$$

D
$$14$$ or $$21$$

Solution

$$\textbf{Step 1: Find the value of n}$$

$$\text{If a,b,c are in A.P,then}$$

$$2b=a+c$$

$$\text{So applying the same conditions, we have}$$

$$\Rightarrow$$ $$2 \times\left({ }^{n} c_{5}\right)=\left({ }^{n} C_{6}\right)+\left({ }^{n} C_{4}\right)$$

$$\text{We know that,}$$

  $${ }^{n} C_{r} = \dfrac{ n !}{r !(n-r) !}$$

$$\Rightarrow$$$$\dfrac{2 \times n !}{5 !(n-5) !}=\dfrac{n !}{6 !(n-6) !}+\dfrac{n !}{4 !(n-4) !}$$

$$\Rightarrow \dfrac{2}{5 !(n-5) !}=\dfrac{1}{6(n-6) !}+\dfrac{1}{4 !(n-4) !} $$

$$\Rightarrow \dfrac{2}{5 !(n-5)(n-6) !}=\dfrac{1}{6 !(n-6) !}+\dfrac{1}{4 !(n-4)(n-5)(n-6)!} $$

$$ \Rightarrow \dfrac{2}{5 !(n-5)}=\dfrac{1}{6 !}+\dfrac{1}{4 !(n-4)(n-5)} $$

$$\Rightarrow \dfrac{2}{5 \times 4 ! \times(n-5)}=\dfrac{1}{6 \times 5 \times 4 !}+\dfrac{1}{4 !(n-4)(n-5)} $$

$$=\dfrac{2}{5 \times(n-5)}=\dfrac{1}{6 \times 5}+\dfrac{1}{(n-4)(n-5)}$$

$$\Rightarrow \dfrac{2}{5 \times(n-5)}-\dfrac{1}{(n-4)(n-5)}=\dfrac{1}{6 \times 5}$$

  $$\Rightarrow 12 n-78=n^{2}-9 n+20$$

$$\Rightarrow n^2-21n+98=0$$

$$\Rightarrow (n-7)(n-14)=0$$

$$\therefore n=14$$ $$\text{(or)}$$ $$n=7$$

$$\textbf{Hence, option A is correct.}$$
Question 6
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In how many ways can $$8$$ persons consisting of $$4$$ men and $$4$$ women can be seated round a table if no two men are to be in adjacent seats.

A
$$4!\times4!$$

B
$$7!$$

C
$$3!\times4!$$

D
$$8!$$

Solution

$$8$$ Persons
Let the men be seated first
$$=(4-1)!\\ =3!\\ =6$$
Number of place for women $$=4$$
$$4$$ women are there
Therefore number of ways $$=4!$$
Total ways $$=4!\times 3!$$
Question 7
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If $$^nC_3=^nC_{13}$$, then $$^{20}C_n$$ is.

A
$$1825$$

B
$$3801$$

C
$$4845$$

D
$$300$$

Solution

We know that $$ { { n }_{ C } }_{ r }= { { n }_{ C } }_{n- r }$$

Given, $$ { { n }_{ C } }_{3}= { { n }_{ C } }_{13} $$
$$ => r = 3 $$ and $$ n -r = 13 $$
$$ => n - 3 = 13 $$
$$ => n = 16 $$

Also, $$ { { n }_{ C } }_{ r }=\dfrac { n! }{ r!(n-r)! } $$
So, $$ { { 20 }_{ C } }_{ 16 }=\dfrac { 20! }{ 16!(20-16)! } = \dfrac { 20! }{ 16! \times 4! } = \dfrac { 20 \times 19 \times 18 \times 17 \times 16! }{ 16! \times 4 \times 3 \times 2 } = 4845 $$
Question 8
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If $$^{19}C_r$$ and $$^{19}C_{r-1}$$ are in the ratio 2:3, then find $$^{14}C_r$$

A
91

B
81

C
71

D
61

Solution

We know that, $$ { { n }_{ C } }_{ r }=\dfrac { n! }{ r!(n-r)! } $$

Given, $$ \dfrac { { { 19 }_{ C } }_{ r } }{ { { 19 }_{ C } }_{ r-1 } } =\quad \dfrac { 2 }{ 3 } $$

$$ =>\dfrac { \dfrac { 19! }{ r!(19-r)! } }{ \dfrac { 19! }{ (r-1)!(19-r+1)! } } =\quad \dfrac { 2 }{ 3 } $$

$$ => \dfrac { (r-1)!(20-r)! }{ r!(19-r)! } =\quad \dfrac { 2 }{ 3 } $$

$$ =\dfrac { (r-1)!\times (20-r) \times (19-r)! }{ r\quad \times (r-1)!\times (19-r)! } =\quad \dfrac { 2 }{ 3 } $$
$$ => \dfrac { 20-r }{ r\quad } =\quad \dfrac { 2 }{ 3 } $$
$$ => r = 12 $$

So,
$$ { { 14 }_{ C } }_{ 12 }=\dfrac { 14! }{ 12!(14-12)! } =\dfrac { 14! }{ 12!(2)! } =\dfrac { 14\times 13\times 12! }{ 12!\times 2 } =\quad 7\times 13=91 $$
Question 9
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If j, k, and n are consecutive integers such that $$0 < j < k < n$$ and the units (ones) digit of the product jn is 9, what is the units digit of k ?

A
0

B
1

C
2

D
3

E
4

Solution

There are only a few ways you can make a product have a units digit of $$9$$. Either both numbers you’re multiplying have to be $$3$$, or one has to be $$1$$ and one has to be $$9$$. Since we’re dealing with consecutive integers, $$j$$ and $$n$$ can’t both end in $$3$$, so they’re going to have to end in $$1$$ and $$9$$.
It’s important to remember that although we only really care about the units digits in this problem, we’re dealing with numbers that might (or in fact, must) be $$2$$ or more digits. That’s why you can have $$k's$$ units digit be $$0$$.
Say $$j= 19$$, $$k= 20$$, and $$n= 21$$. Then $$jn = (19)(21) = 399$$ (units digit is $$9$$), and the units digit of $$k = 0$$.
Hence option A is correct.
Question 10
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The line $$\displaystyle 3x+2y=24$$ meets x-axis at A and y-axis at B. The perpendicular bisector of $$\displaystyle \overline { AB } $$ meets the line through (0, -1) and parallel to x-axis at C. Find the area of $$\displaystyle \Delta ABC$$.

A
85 $$\displaystyle { units }^{ 2 }$$

B
87 $$\displaystyle { units }^{ 2 }$$

C
90 $$\displaystyle { units }^{ 2 }$$

D
91 $$\displaystyle { units }^{ 2 }$$

Solution

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Permutations and Combinations Test 47

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Permutations and Combinations Test 47

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Question 1

$$\displaystyle 3\sqrt { 3 } $$

$$\displaystyle \frac { 3\sqrt { 3 } }{ 8 } $$

$$\displaystyle 12\sqrt { 3 } $$

$$\displaystyle \frac { 3\sqrt { 3 } }{ 4 } $$

Solution

Question 2

$$11!$$

$$12!$$

$$\dfrac{12!}{2}$$

$$1$$

Solution

Question 3

$$19!$$

$$\dfrac{19!}{10!}$$

$$\dfrac{19!}{10!^2}$$

$$20!$$

Solution

Question 4

$$(-1)^x = 1; (-1)^y = 1$$

$$(-1)^x = 1; (-1)^y = -1$$

$$(-1)^x = -1; (-1)^y = 1$$

$$(-1)^x = -1; (-1)^y = -1$$

Solution

Question 5

$$7$$ or $$14$$

$$7$$

$$14$$

$$14$$ or $$21$$

Solution

Question 6

$$4!\times4!$$

$$7!$$

$$3!\times4!$$

$$8!$$

Solution

Question 7

$$1825$$

$$3801$$

$$4845$$

$$300$$

Solution

Question 8

91

81

71

61

Solution

Question 9

0

1

2

3

4

Solution

Question 10

85 $$\displaystyle { units }^{ 2 }$$

87 $$\displaystyle { units }^{ 2 }$$

90 $$\displaystyle { units }^{ 2 }$$

91 $$\displaystyle { units }^{ 2 }$$

Solution

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