Permutations and Combinations Test 15

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Permutations and Combinations Test 15

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Question 1
1 / -0

$$\displaystyle ^{5}C_{4}=$$___

A
$$4$$

B
$$5$$

C
$$20$$

D
$$24$$

Solution

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

So, $$ ^5C_4 = \dfrac { 5! }{ (4!) (5-4)! } =\dfrac { 5! }{ 4! } = \dfrac {5 \times 4 !}{4 ! } = 5 $$
Question 2
1 / -0

In how many ways can 8 books be distributed among 5 students if each student is eligible for any number of books?

A
$$40$$

B
$$\displaystyle 5^{8}$$

C
$$360$$

D
$$\displaystyle 8^{5}$$

Solution

First book can be distributed to student in 5 ways, 2nd book again by 5 ways.
So, using same logic the total ways $$=5\times5\times5.......8$$ times
$$=5^8$$
Hence, the answer is $$5^8.$$
Question 3
1 / -0

In $$\displaystyle ^{3}C_{r}$$ the value of r can be ___

A
$$2$$

B
$$4$$

C
$$6$$

D
All of these

Solution

A combination is a "one or more elements selected from a set in any order

So, $$ {3}_{{ C }_{ r }} $$ means $$ r $$ elements are selected from $$ 3 $$ elements

This is possible only when $$ r \leq 3 $$
In the given options, $$ r = 2 $$ is the correct answer.
Question 4
1 / -0

In how many ways can 3 diamond cards be drawn simultaneously from a pack of cards?

A
$$78$$

B
$$1716$$

C
$$286$$

D
$$13$$

Solution

Pack of cards $$52$$
Diamond cards are $$13$$
To select $$3$$ cards
Total number of choices =$$^{ 13 }C_{ 3 }$$
$$= \dfrac { 13\times 12\times 11 }{ 3\times 2\times 1 } \\ = 143\times 2\\ = 286$$
Hence $$286$$ ways are possible and is a correct answer.
Question 5
1 / -0

The diagram above shows the various paths along which a mouse can travel from point X, where it is released, to point Y, where it is rewarded with a food pellet. How many different paths from X to Y can the mouse take if it goes directly from X to Y without retracing any point along a path?

A
$$6$$

B
$$7$$

C
$$12$$

D
$$14$$

E
$$17$$

Solution

We can calculate the total number of ways by using the basic principle for counting by multiplication.
This we do by multiplying the total possible routes between every two nodes in the path form X to Y. (different node are named in the image )
X $$\rightarrow$$ A $$\rightarrow$$ C and X $$\rightarrow$$ B $$\rightarrow$$ C ------------- 2 ways
C to D has to be covered to move ahead.
D $$\rightarrow$$ E $$\rightarrow$$ G and D$$\rightarrow$$ F $$\rightarrow$$ G ------------- 2 ways
G to H has to be covered to move ahead.
H $$\rightarrow$$ I $$\rightarrow$$ J and H$$\rightarrow$$ J and H$$\rightarrow$$ K $$\rightarrow$$ J ------------- 3 ways
J to Y must be moved to reach Y.
So, total ways = $$ 2 \times 2 \times 2 \times 3 = 12$$ (option C)
Question 6
1 / -0

A researcher plans to identify each participant in a $$174$$ certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are $$12$$ participants,and each participant is to receive a different code?

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

E
$$8$$

Solution

The number of single letter code possible are $$n$$ and no of distinct codes $$={^n{C}_{2}}$$
So, $${^n{C}_{2}}+n\ge 12$$
$$\dfrac{n(n-1)}{2}+n\ge 12$$
$$n(n+1)\ge 24$$
$$n$$ min $$=5$$ as least number is asked.
Hence the answer is $$5.$$
Question 7
1 / -0

A three-digit code for certain locks uses the digits $$0, 1, 2, 3, 4, 5, 6, 7, 8, 9$$ according to the following constraints. The first digit cannot be $$0$$ or $$1$$, the second digit must be $$0$$ or $$1$$, and the second and third digits cannot both be $$0$$ in the same code. How many different codes are possible?

A
$$144$$

B
$$152$$

C
$$160$$

D
$$168$$

E
$$176$$

Solution

The first digit can be fill in $$8$$ ways. For $$2nd$$ digit it can be one of $$(0,1)$$
Now, assume second digit as $$1$$ the $$3rd$$ digit can take $$10$$ values.
$$\therefore$$ Number of codes $$=8\times1\times9=72$$
$$\therefore$$ total $$80+72=152$$
Hence, the answer is $$152.$$
Question 8
1 / -0

Four speakers will address a meeting where speaker $$Q$$ will always speak after $$P$$. Then, the number of ways in which the order of speakers can be prepared is

A
$$256$$

B
$$128$$

C
$$24$$

D
$$12$$

Solution

Four speakers will address the meeting in $$4!$$ ways $$= 24$$ different ways in which half number of cases will be such that $$P$$ speaks before $$Q$$ and half number of case will be such that $$P$$ speaks after $$Q$$
$$\therefore$$ Required number of ways $$=\dfrac {24}{2} = 12$$
Question 9
1 / -0

Mario's Pizza has 2 choices of crust: deep dish and thin-and-crispy. The restaurant also has a choice of 5 toppings: tomatoes, sausage, peppers, onions, and pepperoni. Finally, Mario's offers every pizza in extra cheese as well as regular. If Linda's volleyball team decides to order a pizza with 4 toppings, how many different choices do the teammates have at Mario's Pizza?

A
$$24$$

B
$$32$$

C
$$28$$

D
$$20$$

Solution

20 choices: Consider the toppings first. Model the toppings with the "word" YYYYN, in which four of the toppings are on the pizza and one is not. The number of anagrams for this "word" is:
$$\dfrac{5!}{4!} = 5$$
If each of these pizzas can also be offered in 2 choices of crust, there are $$5 \times 2 = 10$$ pizzas. The same logic applies for extra cheese and regular: $$10 \times 2 = 20.$$
Question 10
1 / -0

From the consecutive integers -10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A
$$\displaystyle { \left( -10 \right) }^{ 20 }$$

B
$$\displaystyle { \left( -10 \right) }^{ 10 }$$

C
$$\displaystyle 0$$

D
$$\displaystyle { -\left( 10 \right) }^{ 19 }$$

E
$$\displaystyle { -\left( 10 \right) }^{ 20 }$$

Solution

For least value we take$$10;19$$ times $$-10$$ once
So value $$10^{19}\times-10$$
$$=(-10)^{20}$$
Hence the answer is $$(-10)^{20}.$$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 15

Result

Permutations and Combinations Test 15

Score

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SHARING IS CARING

Question 1

$$4$$

$$5$$

$$20$$

$$24$$

Solution

Question 2

$$40$$

$$\displaystyle 5^{8}$$

$$360$$

$$\displaystyle 8^{5}$$

Solution

Question 3

$$2$$

$$4$$

$$6$$

All of these

Solution

Question 4

$$78$$

$$1716$$

$$286$$

$$13$$

Solution

Question 5

$$6$$

$$7$$

$$12$$

$$14$$

$$17$$

Solution

Question 6

$$4$$

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 7

$$144$$

$$152$$

$$160$$

$$168$$

$$176$$

Solution

Question 8

$$256$$

$$128$$

$$24$$

$$12$$

Solution

Question 9

$$24$$

$$32$$

$$28$$

$$20$$

Solution

Question 10

$$\displaystyle { \left( -10 \right) }^{ 20 }$$

$$\displaystyle { \left( -10 \right) }^{ 10 }$$

$$\displaystyle 0$$

$$\displaystyle { -\left( 10 \right) }^{ 19 }$$

$$\displaystyle { -\left( 10 \right) }^{ 20 }$$

Solution

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