Permutations and Combinations Test 24

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Permutations and Combinations Test 24

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Weekly Quiz Competition

Question 1
1 / -0

If $$\left( 2\le r\le n \right) $$, then $${ _{ }^{ n }{ C } }_{ r }+2{ _{ }^{ n }{ C } }_{ r+1 }+{ _{ }^{ n }{ C } }_{ r+2 }$$ is equal to

A
$$2.{ _{ }^{ n+2 }{ C } }_{ r+2 }\quad $$

B
$${ _{ }^{ n+r }{ C } }_{ r+1 }$$

C
$${ _{ }^{ n+2 }{ C } }_{ r+2 }$$

D
$${ _{ }^{ n+1 }{ C } }_{ r }$$

Solution

To solve this, we'll use the relation $$^nC_{r}+^nC_{r+1}=^{n+1}C_{r+1}$$ ...[1]

$$^nC_{r}+2^nC_{r+1}+^nC_{r+2}$$

$$=(^nC_{r}+^nC_{r+1})+(^nC_{r+1}+^nC_{r+2})$$

$$= ^{n+1}C_{r+1}+^{n+1}C_{r+2}$$ ...(Using [1])

$$=^{n+2}C_{r+2}$$ ...(Using [1])
Question 2
1 / -0

From a collection of $$20$$ consecutive natural numbers, four are selected such that they are not consecutive. The number of such selections is

A
$$284\times 17$$

B
$$285\times 17$$

C
$$284\times 16$$

D
$$285\times 16$$

Solution

$$1,2,3,4......20$$

there are $$17$$ way for four consecutive number$$\{(1,2,3,4),(2,3,4,5),(3,4,5,6),\dots \dots ,(17,18,19,20)\}$$

numbers ways $$={ _{ }^{ 20 }{ C } }_{ 4 }-17=285\times 17-17=284\times 17$$
Question 3
1 / -0

The number of selection of $$n$$ objects from $$2n$$ objects of which $$n$$ are identical and the rest are different is

A
$${ 2 }^{ n }$$

B
$${ 2 }^{ n-1 }$$

C
$${ 2 }^{ n }-1\quad $$

D
$${ 2 }^{ n }+1\quad $$

Solution

Let $$S_1$$ be the set of $$n$$ identical objects and $$S_2$$ be the set of the remaining $$n$$ different objects.
Number of ways to select $$r$$ objects from $$S_1=1$$ since all objects in $$S_1$$ are identical.
Number of ways to select $$r$$ objects from $$S_2= ^nC_{r}$$.

Total number of ways to select $$n$$ objects =
$$\bullet$$$$n$$ objects from $$S_2$$ and $$0$$ object from $$S_1$$
Number of ways $$= ^nC_{n}$$
$$\bullet$$$$n-1$$ objects from $$S_2$$ and $$1$$ object from $$S_1$$
Number of ways $$ = ^nC_{n-1}\times 1=^nC_{n-1}$$
$$\bullet$$ $$n-2$$ objects from $$S_2$$ and $$2$$ object from $$S_1$$
Number of ways $$ = ^nC_{n-2}\times 1=^nC_{n-2}$$
$$\bullet$$ $$n-3$$ objects from $$S_2$$ and $$3$$ object from $$S_1$$
Number of ways $$ = ^nC_{n-3}\times 1=^nC_{n-3}$$
$$.$$
$$.$$
$$.$$
$$.$$
$$\bullet$$ $$2$$ objects from $$S_2$$ and $$n-2$$ object from $$S_1$$
Number of ways $$ = ^nC_{2}\times 1=^nC_{2}$$
$$\bullet$$ $$1$$ objects from $$S_2$$ and $$n-1$$ object from $$S_1$$
Number of ways $$ = ^nC_{1}\times 1=^nC_{1}$$
$$\bullet$$ $$0$$ objects from $$S_1$$ and $$n$$ object from $$S_1$$
Number of ways $$ = ^nC_{0}\times 1=^nC_{0}$$

$$\therefore$$ Total number of ways $$= ^nC_{n}+^nC_{n-1}+^nC_{n-2}+^nC_{n-3}+....+^nC_{2}+^nC_{1}+^nC_{0}=2^n$$

So, the answer is option (A).
Question 4
1 / -0

The number of ways in which $$6$$ rings can be worn on the four fingers of one hand is

A
$$4^{6}$$

B
$$^{6}C_{4}$$

C
$$6^{4}$$

D
None of these

Solution

Each ring can be worn on $$4$$ fingers, means each have $$4$$ possibilities.
$$\therefore$$ Total ways$$=4\times4\times4\times4\times4\times4=4^{6}$$
Hence, $$(A)$$
Question 5
1 / -0

Ramesh number of ways in which the letters of the word RAMESH can be placed in the squares of the given figure so that no row remains empty, is

A
17280

B
18720

C
15840

D
14400

Solution

Case 1 $$2.2.2$$ from top to bottom
$$(\ ^{6}C_{2}\times 2!)\times (\ ^{4}C_{2}\times 2!)\times (4\times 3)=4320$$
Case 2 $$2, 1, 3$$ from top the bottom:
$$(\ ^{6}C_{2}\times 2!)\times (\ ^{4}C_{1}\times 2)\times (4\times 3\times 2)=5760$$
Case 3: $$1, 2, 3$$ from top of bottom
$$(\ ^{6}C_{1}\times 2)\times (\ ^{5}C_{2}\times 2!)\times (4\times 3\times 2)=5760$$
Case 4: $$1,1,4$$ from top the bottom
$$(\ ^{6}C_{1}\times 2)\times (\ ^{5}C_{1}\times 2)\times (4!)$$
So the correct answer is $$(B) 18, 720$$
Question 6
1 / -0

Let n be the number of ways in which the letters of the word "RESONANCE" can be arranged so the vowels appear at the even places and m be the number of ways in which "RESONANCE" can arrange so that letters R,S,O,A appears in the order same as the word RESONANCE, then answers the following questions.
The value of n is

A
360

B
720

C
240

D
840

Solution

RESONANCE
VOWELS $$=E, O, A, E$$
CONSTANTS $$=k, S, N, N, C$$
$$\bar{1}\ \bar{(2)}\ \bar{3}\ \bar{(4)}\ \bar{5}\ \bar{(6)}\ \bar{7}\ \bar{(8)}\ \bar{9}$$
$$\Rightarrow$$ vowels can be arranged as $$=\dfrac{4!}{2!}$$
$$\Rightarrow$$ consonants can be as $$=\dfrac{5!}{2!}$$
Total No. of ways
$$\dfrac{4!}{2!}\times \dfrac{5!}{2!}=\dfrac{24}{2}\times \dfrac{120}{2}=12\times 60 =720$$
Question 7
1 / -0

For $$n$$ being natural number, if $$^{2n}C_r={}^{2n}C_{r+2}$$, find $$r$$.

A
$$n$$

B
$$n-1$$

C
$$n-2$$

D
$$n-3$$

Solution

Given,
$$^{2n}C_r=^{2n}C_{r+2}$$
or,
$$^{2n}C_{2n-r}=^{2n}C_{r+2}$$ [ As $$^nC_r=^nc_{n-r}$$]
or,
$$2n-r=r+2$$
or,
$$2r=2n-2$$
or,
$$r=n-1$$.
Question 8
1 / -0

The number of ways in which we can select 5 letters of the word INTERNATIONAL is equal to

A
$$200$$

B
$$220$$

C
$$242$$

D
$$256$$

Solution

We have thirteen letters.
$$2l, 3N, 2T, 2A$$ and $$(E, O, L,R)8$$ (types)
We can select 5(five) letters in the following manner:

1. All different $${^8C}_5 = \dfrac{8.7.5}{1.2.3} = 56$$ ways

2. 2 alike, 3 different $${^4C}_1 . {^7C}_3 = 435 = 140$$ ways (we have 4 sets of alike letters)

3. 3 alike, 2 different $${^1C}_1 . {^7C}_2 = 1.21 = 21$$ ways (we have only one set of 3 alike)

4. 3 alike and 2 alike $${^1C}_1 . {^3C}_1 = 1.3 = 3$$ ways

5.Two sets of alike and one different
$${^4C}_2 . {^6C}_1 = 6.6 = 36$$ ways

$$\therefore$$ Total number of selection is
$$56 + 140 + 21 + 3 + 36 = 256$$ ways $$\Longrightarrow$$ (d)
Question 9
1 / -0

How many different $$4$$-person committees can be chosen form the $$100$$ members of the Senate ?

A
$$25$$

B
$$400$$

C
$$3,921,225$$

D
$$94,109,400$$

Solution

Total number of members $$=100$$.
Members has to be selected $$=4$$.
$$\therefore$$ different committees $$=^{100}C_{4}=\large{\frac{100!}{4!\times96!}}$$ $$=3,921,225$$.
Question 10
1 / -0

If the letterss of the word 'NAAGI' are arranged as in a dictionary then the rank of the given word is

A
$$23$$

B
$$84$$

C
$$49$$

D
$$48$$

Solution

In a dictionary, letters in alphabetical order are $$A,A,G,I,N$$

so the first word will be $$AAGIN$$

Number of words beginning with $$A=$$ Number of ways arranging $$A,G,I,N$$$$=4!=24$$
Number of words beginning with $$G=\dfrac{4!}{2!}=12$$

Number of words beginning with $$I=\dfrac{4!}{2!}=12$$

Next word will be $$NAAGI$$

No. of words before $$NAAGI$$ $$=24+12+12=48$$

So rank of the word $$NAAGI$$ is $$49$$.

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Permutations and Combinations Test 24

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Permutations and Combinations Test 24

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Question 1

$$2.{ _{ }^{ n+2 }{ C } }_{ r+2 }\quad $$

$${ _{ }^{ n+r }{ C } }_{ r+1 }$$

$${ _{ }^{ n+2 }{ C } }_{ r+2 }$$

$${ _{ }^{ n+1 }{ C } }_{ r }$$

Solution

Question 2

$$284\times 17$$

$$285\times 17$$

$$284\times 16$$

$$285\times 16$$

Solution

Question 3

$${ 2 }^{ n }$$

$${ 2 }^{ n-1 }$$

$${ 2 }^{ n }-1\quad $$

$${ 2 }^{ n }+1\quad $$

Solution

Question 4

$$4^{6}$$

$$^{6}C_{4}$$

$$6^{4}$$

None of these

Solution

Question 5

17280

18720

15840

14400

Solution

Question 6

360

720

240

840

Solution

Question 7

$$n$$

$$n-1$$

$$n-2$$

$$n-3$$

Solution

Question 8

$$200$$

$$220$$

$$242$$

$$256$$

Solution

Question 9

$$25$$

$$400$$

$$3,921,225$$

$$94,109,400$$

Solution

Question 10

$$23$$

$$84$$

$$49$$

$$48$$

Solution

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