Permutations and Combinations Test 13

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Permutations and Combinations Test 13

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Weekly Quiz Competition

Question 1
1 / -0

If $$^nC_3 + ^nC_4 > ^{n+1}C_3$$, then

A
$$n > 6$$

B
$$n > 7$$

C
$$n < 6$$

D
none of these

Solution

$$\:^{n}C_{3}+\:^{n}C_{4}$$
$$=\:^{n+1}C_{4}$$
$$=\:^{n+1}C_{n-3}$$
Now
$$\:^{n+1}C_{n-3}>\:^{n+1}C_{3}$$
Hence
$$n-3>3$$
$$n>6$$.
Question 2
1 / -0

If $$4.^nC_6 = 33.^{n-3}C_3$$ then $$n$$ is equal to

A
$$9$$

B
$$10$$

C
$$11$$

D
none of these

Solution

$$4.\:^{n}C_{6}=33.(\:^{n-3}C_{3})$$
$$4.(\dfrac{n!}{(n-6)!.6!})=33.(\dfrac{(n-3)!}{3!.(n-6)!}$$
$$4(n(n-1)(n-2))=33(4\times5\times6)$$
$$n(n-1)(n-2)=11\times3\times5\times6=11.10.9$$

Hence the real root that we get is
$$n=11$$.
Hence $$n=11$$.
Question 3
1 / -0

If $$2\times$$ $$^nC_5 = 9\times$$ $$^{n-2}C_5$$, then the value of n will be:

A
7

B
10

C
9

D
5

Solution

Given, $$2(\:^{n}C_{5})=9(\:^{n-2}C_{5})$$
$$2.\dfrac{n!}{5!.(n-5)!}=9.\dfrac{(n-2)!}{(n-7)!.5!}$$

$$\dfrac{2(n-1)(n)}{(n-6)(n-5)}=9$$
$$2n^{2}-2n=9(n^{2}-11n+30)$$
$$7n^{2}-97n+270=0$$
Hence $$n=10$$ and $$n=\dfrac{27}{7}$$
Since $$n\epsilon N$$.
Hence $$n=10$$.
Question 4
1 / -0

$$\displaystyle \sum_{r=0}^m {\;}^{n+r}C_n=$$

A
$$^{n+m+1}C_{n+1}$$

B
$$^{n+m+2}C_{n}$$

C
$$^{n+m+3}C_{n+1}$$

D
none of these

Solution

Given $$\sum _{ r=0 }^{ m } {^{ n+r }C_{ n }}$$
$$=^{ n }C_{ n }+^{ n+1 }C_{ n }+^{ n+2 }C_{ n }+^{ n+3 }C_{ n }.....^{ n+m }C_{ n }$$

$$=^{ n+1 }C_{ n+1 }+^{ n+1 }C_{ n }+^{ n+2 }C_{ n }+^{ n+3 }C_{ n }.....^{ n+m }C_{ n }$$ ($$\because ^{ n }C_{ n }=^{ n+1 }C_{ n+1 }$$)

$$=^{ n+2 }C_{ n+1 }+^{ n+2 }C_{ n }+^{ n+3 }C_{ n }.....^{ n+m }C_{ n }$$
($$\because ^nC_r+^nC_{r-1}=^{n+1}C_r$$)

$$=^{ n+3 }C_{ n+1 }+^{ n+3 }C_{ n }+......^{ n+m }C_{ n }$$
Continuing this ,
$$=^{ n+m }C_{ n+1 }+^{ n+m }C_{ n }$$
$$=^{ n+m+1 }C_{ n+1 }$$
Question 5
1 / -0

If $$^{15}C_{3r}=^{15}C_{r+3}$$, then the value of r is:

A
3

B
4

C
5

D
8

Solution

Since, $$^nC_r = ^nC_{n-r}$$
Either $$ \Longrightarrow 3r=r+3$$ or $$ 3r=15-(r+3)$$
Hence, $$ r=\dfrac { 3 }{ 2 } $$ or $$r=3$$
Since r is an integer, therefore $$r=3 $$
Question 6
1 / -0

Let $$A=(x|x$$ is a prime number and $$x<300$$ > the number of different rational numbers, whose numerator and denominator belong to $$A$$ is:

A
91

B
84

C
106

D
None of these

Solution

Set A contains 10 elements. 2 different numbers for numerator and denominator from these can be obtained in $$10\times 9=90$$ ways and each permutation will form a unique rational number different from one. In addition, one will be formed when numerator and denominator are same. So required number of numbers $$90+1=91$$
Question 7
1 / -0

If n and r are two positive integers such that $$n\geq r$$, then $$^nC_{r+1} + ^nC_r =$$

A
$$^nC_{n-r}$$

B
$$^nC_r $$

C
$$^{n-1}C_r $$

D
$$^{n+1}C_r+1 $$

Solution

The value of $$\:^{n}C_{r+1}+\:^{n}C_{r}$$

$$=\dfrac{n!}{(n-(r+1))!(r+1)!}+\dfrac{n!}{(n-r)!.r!}$$

$$=\dfrac{n!}{r!.(n-(r+1))!}[\dfrac{1}{r+1}+\dfrac{1}{n-r}]$$

$$=\dfrac{n!}{r!.(n-(r+1))!}[\dfrac{n+1}{(r+1)(n-r)}]$$

$$=\dfrac{(n+1)!}{(n-r)!(r+1)!}$$

$$=\:^{n+1}C_{r+1}$$.
Question 8
1 / -0

If $$(^{15}C_r + ^{15}C_{r-1}) (^{15}C_{15-r} + ^{15}C_{16-r}) = (^{16}C_{13})^2$$, then the value of $$r$$ is

A
$$r=2$$

B
$$r=3$$

C
$$r=4$$

D
none of these

Solution

$$(^{ 15 }C_{ r }+^{ 15 }C_{ r-1 })(^{ 15 }C_{ 15-r }+^{ 15 }C_{ 16-r })=(^{ 16 }C_{ 13 })^{ 2 }$$
$$^{ 16 }C_{ r }(^{ 15 }C_{ r }+^{ 15 }C_{ r-1 })=(^{ 16 }C_{ 13 })^{ 2 }$$ ($$\because ^nC_r+^nC_{r-1}=^{n+1}C_r$$)
$$^{ 16 }C_{ r }(^{ 16 }C_{ r })=(^{ 16 }C_{ 3 })^{ 2 }$$ ($$\because ^nC_r=^nC_{n-r}$$)
$$(^{ 16 }C_{ r })^{ 2 }=(^{ 16 }C_{ 3 })^{ 2 }$$
$$\Rightarrow r=3$$
Question 9
1 / -0

In a game called 'odd man out', $$ m (m > 2)$$ persons toss a coin to determine who will buy refreshment for the entire group. A person who gets an outcome different from that of the rest of the members of the group is called the odd man out. The probability that there is a loser in any game is

A
$$1/2 m$$

B
$$m/2 ^{m-1}$$

C
$$2/m$$

D
none of these

Solution

Let us represent the person as $${ P }_{ 1 }$$ , $${ P }_{ 2 }$$ , $${ P }_{ 3 }$$ , $${ P m }.$$
For each person we have 2 outcome,
So, total outcomes $$= { 2 }^{ m }$$
Favorable outcomes:
$$1.$$ One gets tail rest $$\rightarrow$$ Head
$$2.$$ One get head test $$\rightarrow$$ Tail
No. of ways $$m\times 1$$ ($$m=$$ because in people).
Also for second condition we have $$m$$ ways.
Total favorable ways $$=\dfrac { 2m }{ { 20 }^{ m } } ={ m }/{ { 2 }^{ m-1 } }$$
Hence, answer is $${ m }/{ { 2 }^{ m-1 } }.$$
Question 10
1 / -0

Three players play a total of $$9$$ games. In each game, one person wins and the other two lose; the winner gets $$2$$ points and the losers lose $$1$$ each. The number of ways in which they can play all the $$9$$ games and finish each with a zero score is

A
$$84$$

B
$$1680$$

C
$$7056$$

D
$$0$$

Solution

Each player need to win $$3$$ matches and lose $$6$$ matches to get a zero score.
Out of $$9$$ games $$3$$ should be selected by first player(games which he will win)$$=^9C_3$$
Out of $$6$$ games 3 should be selected by second player $$={}^6C_3$$
Out of $$3$$ games $$3$$ should be selected by third player $$={}^3C_3$$
Hence the answer is $$={}^9C_3\times^6C_3\times^3C_3=1680$$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 13

Result

Permutations and Combinations Test 13

Score

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Rank

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SHARING IS CARING

Question 1

$$n > 6$$

$$n > 7$$

$$n < 6$$

none of these

Solution

Question 2

$$9$$

$$10$$

$$11$$

none of these

Solution

Question 3

7

10

9

5

Solution

Question 4

$$^{n+m+1}C_{n+1}$$

$$^{n+m+2}C_{n}$$

$$^{n+m+3}C_{n+1}$$

none of these

Solution

Question 5

3

4

5

8

Solution

Question 6

91

84

106

None of these

Solution

Question 7

$$^nC_{n-r}$$

$$^nC_r $$

$$^{n-1}C_r $$

$$^{n+1}C_r+1 $$

Solution

Question 8

$$r=2$$

$$r=3$$

$$r=4$$

none of these

Solution

Question 9

$$1/2 m$$

$$m/2 ^{m-1}$$

$$2/m$$

none of these

Solution

Question 10

$$84$$

$$1680$$

$$7056$$

$$0$$

Solution

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