Permutations and Combinations Test 52

Result

Permutations and Combinations Test 52

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Weekly Quiz Competition

Question 1
1 / -0

If $$(1+x)^{15}=C_{0}+C_{1}xC_{2}x^{2}+...+C_{15}x^{15},$$ then $$^{15}C_{0}^{2}- ^{15}C_{1}^{2}+^{15}C_{2}^{2}- ^{15}C_{2}^{3}+... ^{15}C_{15}^{2}$$ is equal to

A
0

B
1

C
-1

D
none of these

Solution
Question 2
1 / -0

How many different $$5$$ letter sequences can be made using the letters $$A,B,C,D$$ with repetition such that the sequence does not include the word $$BAD$$?

A
$$1024$$

B
$$976$$

C
$$48$$

D
$$678$$

Solution
Question 3
1 / -0

If $$^{n}C_{1}=^{n}C_{2}$$, then $$n$$ is equal to

A
$$2$$

B
$$3$$

C
$$5$$

D
$$none\ of\ these$$

Solution
Question 4
1 / -0

$${ C }_{ 0 }+({ C }_{ 0 }+{ C }_{ 1 })+({ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 })+......+({ C }_{ 0 }+{ C }_{ 1 }+........+{ C }_{ n })=$$

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

B
$$(n+2).{ 2 }^{ n }$$

C
$${ 2 }^{ n }$$

D
$$(n+2).{ 2 }^{ n-1 }$$

Solution
Question 5
1 / -0

If $$^{n}C_{r}=\dfrac {n\ !}{r\ !(n-r)\ !}$$ then the sum of the series $$1+^{n}C_{1}+^{n+1}C_{2}+^{n+2}C_{3}+.....+^{n+r-1}C_{r}$$ is equal to

A
$$^{n+r}C_{r}$$

B
$$^{n+r-1}C_{r}$$

C
$$^{n+r}C_{r-1}$$

D
$$None\ of\ these$$

Solution
Question 6
1 / -0

If the second term of the expansion $$\left[a^{1/13}+\dfrac{a}{\sqrt{a^{-1}}}\right]^n$$ is $$14a^{5/2}$$ then the value of $$\dfrac{^{n}C_3}{^{n}C_2}$$ is

A
$$4$$

B
$$3$$

C
$$12$$

D
$$6$$

Solution

2nd Term $$= ^nC_1(a^{1/13})^{(n-1)} \left ( \frac{a}{\sqrt{a-1}} \right )^1=14 a^{5/2}$$

$$\Rightarrow n\,\,\, a^{\tfrac{n-1}{13}}.a^{3/2}= 14a^{5/2}$$

$$\Rightarrow $$ Comparing both sides

$$\Rightarrow \dfrac{n-1}{13}+\dfrac{3}{2}=\dfrac{5}{2}$$

$$\Rightarrow \dfrac{n-1}{13}=1\Rightarrow n-1=13$$

$$\Rightarrow n=14$$

$$\dfrac{^nC_3}{^nC_2}\Rightarrow \dfrac{\dfrac{n!}{3!(n-3)!}}{\dfrac{n!}{2!(n-2)!}}=\dfrac {2!(n-2)!n!} {n!3!(n-3)!}$$

$$\Rightarrow \dfrac {2!12!}{3! 11!}=\dfrac {12}{3}=4$$
Question 7
1 / -0

If $$\dfrac{c_0}{2}-\dfrac{c_1}{3}+\dfrac{c_2}{4}-.....+(-1)^n\dfrac{c_n}{n+2}=\dfrac{1}{2013\times2014}$$ then n =

A
$$2013$$

B
$$2014$$

C
$$2012$$

D
$$2011$$
Question 8
1 / -0

$$\sum _{ r=0 }^{ n }{ r\left( n-r \right) \left( _{ }^{ n }{ { C }_{ r } } \right) ^{ 2 } } $$ is equal to

A
$$n^{2}._{ }^{ 2n-1 }{ C }_{ n } $$

B
$$n^{2}._{ }^{ 2n }{ { C }_{ n-1 } }$$

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

D
None of these

Solution
Question 9
1 / -0

The value $$\sum _{ r=0 }^{ 10 }{ r_{ }^{ 10 }{ C_{ r } } } .{ \left( -2 \right) }^{ 10-r }$$ is

A
10

B
20

C
30

D
300

Solution
Question 10
1 / -0

If $$^{50}C_{r}=C_{r}$$ then the value of $$C_{0}^{2}+\frac{1}{2}C_{1}^{2}+\frac{1}{3}C_{2}^{2}+\frac{1}{4}C_{3}^{2+....+\frac{1}{50}^{101}C_{51}}$$

A
$$\frac{1}{51}^{101}C_{50}$$

B
$$\frac{1}{50}^{101}C_{50}$$

C
$$\frac{1}{51}^{101}C_{49}$$

D
$$\frac{1}{50}^{101}C_{51}$$

Solution