Permutations and Combinations Test 29

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Permutations and Combinations Test 29

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Weekly Quiz Competition

Question 1
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The number of intersection points of diagonals of $$2009$$ sides polygon, which lie inside the polygon.

A
$$^{2009}C_4$$

B
$$^{2009}C_2$$

C
$$^{2008}C_4$$

D
$$^{2008}C_2$$

Solution

Choosing any $$4$$ vertices make $$1$$ intersection

$$\therefore \ $$ No. of intersection points of diagonals $$=^nC_4$$

Here $$n=2009$$, No. of points $$=^{2009}C_4$$

$$\therefore \ $$ Option $$A$$ is correct.
Question 2
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Number of five-digit numbers divisible by 5 that can be formed from the digits $$0,1, 2, 3, 4, 5$$ without repetition of digits are

A
$$240$$

B
$$360$$

C
$$148$$

D
$$216$$

Solution
Question 3
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Ten persons, amongst whom are $$A$$,$$B$$ and $$C$$ to speak at a function. The number of ways in which it can be done if $$A$$ wants to speak before $$B$$ AND $$B$$ wants to speak before $$C$$ is

A
$$\dfrac{{10!}}{6}$$

B
$$3!$$ $$7!$$

C
$$^{10}{P_3}.7!$$

D
none of these

Solution
Question 4
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Let $$p=\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{5}{1\times6}+.......+\dfrac{1}{2013\times2014}$$ and $$Q=\dfrac{1}{1008\times2014}+\dfrac{1}{1009\times2013}+.........+\dfrac{1}{2014\times1008}$$
then $$\dfrac{P}{Q}=$$

A
$$2013$$

B
$$2014$$

C
$$1511$$

D
$$2$$
Question 5
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The number of ways in which a mixed doubles tennis game can be arranged between 10 players consisting of 6 men and 4 women is .

A
180

B
90

C
48

D
12

Solution

Number of ways to select two men out of six men
$$ = ^{6}C_{2} $$
Number of ways to select two women out of four
women $$ = ^{4}C_{2} $$
Number of ways of selecting the players for the
mixed double $$ = ^{6}C_{2}\times ^{4}C_{2} = 15\times 16 = 90 $$
Let $$ M_{1},M_{2},W_{1}\& W_{2} $$ are selected players for we
mixed double tennis game
if $$ M_{1} $$ chooses $$W_{1}$$ then $$ M_{2}$$ has $$W_{2}$$ as the partner or if
$$ M_{1}$$ chooses $$ W_{2}$$, lean $$M_{2}$$ has $$ W_{1}$$ as the partner
$$ \therefore $$ There are $$2$$ choice for the teams.
Thus, Number of ways in which mixed double
tennis game be arranged $$ 90\times 2 = 180 $$
Question 6
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The number of permutation of the letters of the word $$HINDUSTAN$$ such that neither the pattern $$'HIN'$$ nor $$'DUS'$$ nor $$'TAN'$$ appears, are :

A
$$166674$$

B
$$169194$$

C
$$166680$$

D
$$181434$$

Solution

Total number of letters $$=9$$ in which N is repeated twice

Therefore total number of permutation $$=\dfrac{9!}{2}$$

The number of permutation in which HIN comes as block $$=9-3+1=7!$$

The number of permutation in which DUS comes as block $$=7!$$

The number of permutation in which TAN comes as block $$=\dfrac{7!}{2}$$

The number of permutation in which HIN and DUS comes as block $$=5!$$

The number of permutation in which DUS and TAN comes as block $$=5!$$

The number of permutation in which HIN and TAN comes as block $$=5!$$

Therefore the required number of permutations $$=\dfrac{9!}{2}-\left(7!+7!+\dfrac{7!}{2}-3\times5!+3!\right)$$

$$=\dfrac{362880}{2}-\left(5040+5040+\dfrac{5040}{2}-360+6\right)$$

$$=181440-\left(10080+2520-360+6\right)$$

$$=181440-12246$$

$$=169194$$
Question 7
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The number of ways in which the letters of the word $$"ARRANGE"$$ can be permuted such that $$R's$$ occur together is

A
$$\dfrac{!7}{!2!2}$$

B
$$6!$$

C
$$\dfrac{6!}{2!}$$

D
none of these

Solution
Question 8
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Let the eleven letters, $$A, B, ....K$$ denote an artbitrary permutation of the integers $$(1,2,....11)$$, then $$(A-1)(B-2)(C-3)...(K-11)$$ is

A
Necessarily zero

B
Always odd

C
Always evem

D
None of these

Solution

Given set of numbers is {1,2,3.....11} in which 5 are even and six are odd, which demands that in the given product it is not possible to arrange to subtract only even number from odd numbers. There must be at least one factor involving subtraction of an odd number from another odd number. So at least one of the factors is even. Hence product is always even.
Question 9
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If repetitions are not allowed, the number of numbers consisting of $$4$$ digits and divisible by $$5$$ and formed out of $$0,1,2,3,4,5,6$$ is

A
$$220$$

B
$$240$$

C
$$370$$

D
$$588$$

Solution

$$\rightarrow$$ For divisibility by $$5$$, unit's place should be $$0$$ or $$5$$
with $$0$$, ----$$0$$
$$6$$ choices $$\Rightarrow ^6C_3\times 3!$$
$$=\dfrac{6!}{3!3!}\times 3!=6.5.4$$
$$=120$$
with $$5$$, ----$$5$$
$$6$$ choices $$=120$$
$$^5C_2\times 2!=10\times 2=20$$
$$\Rightarrow 120+120-20=220$$
$$\rightarrow (A)$$
Question 10
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Consider all permutations of the letters of the word MORADABAD.
The number of permutations which contain the word BAD is:

A
$$21 \times 5!$$

B
$$7 \times 5!$$

C
$$6 \times 5!$$

D
$$2 \times 5!$$

Solution

Total number of letters = $$9$$

Consider BAD as a group, then the number of total letters = $$6+1=7$$

$$\therefore$$ Permutation of 7 things = $$7!$$

but letters remaining after removing BAD will have two A's

$$\therefore$$ Total permutation = $$\dfrac{7!}{2}$$ = $$21\cdot 5!$$