Permutations and Combinations Test 17

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Permutations and Combinations Test 17

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Weekly Quiz Competition

Question 1
1 / -0

Kathy is scheduling the first four periods of her school day. She needs to fill those periods with calculus, art, literature, and physics, and each of these courses is offered during each of the first four periods. Calculate the total number of different schedules Kathy can choose from.

A
$$1$$

B
$$4$$

C
$$12$$

D
$$24$$

E
$$120$$

Solution

Number of options of selecting a course for $$1^{st}$$ period is $$4$$(there are $$4$$ courses )
Number of options of selecting a course for $$2^{nd}$$ period is $$3$$
Number of options of selecting a course for $$3^{rd}$$ period is $$2$$
Therefore total number of different schedules is $$4 \times 3 \times 2 \times 1 = 24$$.
Question 2
1 / -0

If $$^{(n-1)}C_3+^{(n-1)}C_4 > ^nC_3$$, then the minimum value of $$n$$ is:

A
5

B
6

C
7

D
8

Solution

$$^{(n-1)}C_3+^{(n-1)}C_4 > ^nC_3$$,
Use $$^nC_{r-1}+^nC_{r}=^{n+1}C_r$$
$$\Rightarrow\dfrac{n!}{(n-4)!4!}>\dfrac{n!}{(n-3)!3!}$$
$$\Rightarrow \dfrac{1}{4}>\dfrac{1}{n-3}$$
$$\Rightarrow n-3>4\Rightarrow n>7$$
Smallest value of $$n=8$$
Question 3
1 / -0

In how many different orders can five boys stand on a line?

A
40

B
50

C
80

D
120

Solution

Five boys can stand in a line in $$5! = 120$$ ways.
Question 4
1 / -0

Six points were chosen on a circle and every possible chord was drawn. Two chords, which do not have the common points are named separately. How many pairs of separate chords exist in the situation described above?

A
$$26$$

B
$$28$$

C
$$30$$

D
$$34$$

Solution

We have to choose six points on circle and join them for every possible chord
So total number of ways is $$6!$$
But Two chord, which do not have common points i.e different chord
Hence Two chord has $$4$$ points So
Number of ways of different or separately chord is $$\dfrac{6!}{4!}=30$$
Question 5
1 / -0

Let $$\boxed { n }$$ be defined as $$\frac{(n+2)!}{(n-1)!}$$, what is the value of $$\frac{\boxed{7}}{\boxed {3}}$$ ?

A
4.4

B
8.4

C
12.4

D
16.4

E
20.4

Solution

Given, box $$n$$ is equal to $$\dfrac{(n+2)!}{(n-1)!}=(n+2)(n+1)n$$
We get box $$7$$ is equal to $$9 \times 8 \times 7 =504$$
We get box $$3$$ is equal to $$5 \times 4 \times 3 = 60$$
Therefore, If we divide those two, we get $$\dfrac{504}{60} = 8.4$$.
Question 6
1 / -0

If $$^6P_r = 360$$ and $$^6C_r =15$$,then find $$r?$$

A
$$5$$

B
$$6$$

C
$$4$$

D
$$3$$

Solution

$$\dfrac{6!}{(6-r)!}=360$$
$$\dfrac{6!}{(6-r)r!}=15$$
$$r!=\dfrac{360}{15}=24$$
$$r=4$$
Question 7
1 / -0

The total number of five digit numbers the sum of whose digits is odd is

A
$$9\times 10^4$$

B
$$\dfrac{9\times 10^4}{2}$$

C
$$10^5$$

D
none of these

Solution

$$0,1,2,3,4,5,6,7,8,9$$
$$5$$ Digit number
If sum of first $$4$$ digits is even then last digit should be odd and if it is odd then it should be even $$=5$$ Ways
Total numbers
$$=9\times 10\times 10\times 10\times 5\\ =4.5\times 10^{ 4 }\\ =\dfrac { 9 }{ 2 } \times 10^{ 4 }$$
Hence the correct answer is $$\dfrac{ 9 }{ 2 }\times 10^{ 4 }$$
Question 8
1 / -0

Find the coefficient of the middle term of the expansion $$\left (x-\dfrac{1}{2y}\right)^{10}$$:

A
$$-\dfrac{63}{8}$$

B
$$24$$

C
$$-\dfrac{33}{4}$$

D
$$-33$$

Solution

We need to find middle term of $$\left (x-\dfrac {1}{2y}\right)^{10}$$
Total number of terms are $$11$$.
So, middle term will be the $$6^{th}$$ term.
$$\therefore t_{5+1}= ^{10}C_5 (-1)^5 x^{10-5} (\dfrac {1}{2y})^5=-\dfrac{63x^5}{8y^5}$$
Hence, option A is correct.
Question 9
1 / -0

A company has $$5$$ men and $$6$$ women. What are the number of ways of selecting a group of eight persons?

A
$$165$$

B
$$185$$

C
$$205$$

D
$$225$$

Solution

The number of ways of selecting a group of eight is
5 men and 3 women=$$^5C_5 \times ^6C_3$$ =20
4 men and 4 women=$$^5C_4 \times ^6C_4$$ =75
3 men and 5 women=$$^5C_3 \times ^6C_5$$=60
2 men and 6 women=$$^5C_2 \times ^6C_6$$=10
Thus the total possible cases is 20+75+60+10=165.
Option A is correct answer.
Question 10
1 / -0

The value of $${ }^{10}C_1 +{ }^{10}C_2 + { }^{10}C_3 + ... + { }^{10}C_9$$ is

A
$$2^{10}$$

B
$$2^{11}$$

C
$$2^{10}-2$$

D
$$2^{10}-1$$

Solution

Consider $$(1+x)^{10}={ }^{10}C_0x^{10}+{ }^{10}C_1x^9+{ }^{10}C_2x^8+.........+{ }^{10}C_9x+{ }^{10}C_{10}$$

Now substitute $$x=1$$ both sides

$$(1+1)^{10}={ }^{10}C_0+{ }^{10}C_1+{ }^{10}C_2+.........+{ }^{10}C_9+{ }^{10}C_{10}$$

$$\Rightarrow 2^{10}=1+{ }^{10}C_1+{ }^{10}C_2+.........+{ }^{10}C_9+1$$

$$\therefore { }^{10}C_1+{ }^{10}C_2+{ }^{10}C_3+......+{ }^{10}C_9=2{ }^{10}-2$$

Note: $${ }^{n}C_0=1$$ and $${ }^nC_n=1$$

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Permutations and Combinations Test 17

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Permutations and Combinations Test 17

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Question 1

$$1$$

$$4$$

$$12$$

$$24$$

$$120$$

Solution

Question 2

5

6

7

8

Solution

Question 3

40

50

80

120

Solution

Question 4

$$26$$

$$28$$

$$30$$

$$34$$

Solution

Question 5

4.4

8.4

12.4

16.4

20.4

Solution

Question 6

$$5$$

$$6$$

$$4$$

$$3$$

Solution

Question 7

$$9\times 10^4$$

$$\dfrac{9\times 10^4}{2}$$

$$10^5$$

none of these

Solution

Question 8

$$-\dfrac{63}{8}$$

$$24$$

$$-\dfrac{33}{4}$$

$$-33$$

Solution

Question 9

$$165$$

$$185$$

$$205$$

$$225$$

Solution

Question 10

$$2^{10}$$

$$2^{11}$$

$$2^{10}-2$$

$$2^{10}-1$$

Solution

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