Permutations and Combinations Test 50

Result

Permutations and Combinations Test 50

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Weekly Quiz Competition

Question 1
1 / -0

$${ if }^{ n }{ c }_{ r }{ + }^{ n }{ c }_{ r+1 }={ n+1 }_{ { C }_{ x } }$$ then the value of x is equal to

A
r

B
r + 1

C
r - 1

D
2r
Question 2
1 / -0

The value of $$\sum _{ r=0 }^{ n }{ \sum _{ s=0 }^{ n }{ (r+s){ C }_{ r }{ C }_{ s } } } ,$$ is

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

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

C
$$n(n+1)2^n$$

D
$$2^{2n}$$

Solution
Question 3
1 / -0

If $$n\ \in\ N$$ & $$n$$ is even, then $$\dfrac {1}{1\ .\ (n-1)\ !}+\dfrac {1}{3\ !(n-3)\ !}+\dfrac {1}{5\ !\ (n-5)\ !}+....+\dfrac {1}{(n-1)\ !\ 1\ !}=$$

A
$$2^{n}$$

B
$$\dfrac {2^{n-1}}{n\ !}$$

C
$$2^{n}n\ !$$

D
$$none\ of\ these$$

Solution
Question 4
1 / -0

$${ if } ^ { n }{ C }_{ 3 }{ + }^{ n }{ C }_{ 4 }>{ n+1 }_{ C_{ 3 } }$$ then

A
n > 6

B
n > 7

C
n < 6

D
none of these
Question 5
1 / -0

if$$^{ 2017 }{ c }_{ 0 }{ + }^{ 2017 }{ c }_{ 1 }{ + }^{ 2017 }{ c }_{ 2 }+.....{ + }^{ 2017 }{ c }_{ 1008 }={ \lambda }^{ 2 }(\lambda >0),$$ then remainder when $$\lambda$$ is divided by 33 is

A
8

B
13

C
17

D
25

Solution

Given $$^{2017}C_{0}+\, ^{2017}C_{1}+\, ^{2017}C_{2}+….+\, ^{2017}C_{1008}= \lambda^{2}$$ $$(\lambda > 0)$$
$$\Rightarrow \displaystyle \sum^{1008}_{k=0}$$ $$^{2017}C_{k}= \lambda^{2}$$
He we know that,
$$\displaystyle \sum^{n}_{k=0}$$ $$^{n}C_{k}=2^{n}$$
$$\displaystyle \therefore \sum_{k=0}^{2017}$$ $$^{2017}C_{k}=2^{2017}$$
$$\Rightarrow \displaystyle \sum_{k=0}^{1008}$$ $$ ^{2017}C_k$$ $$+\sum_{k=1009}^{2017}$$ $$^{2017}C_{k}=2^{2017}$$
$$\Rightarrow \displaystyle \sum_{k}^{1018}$$ $$^{2017}C_{k}+$$ $$^{2017}C_{1009}$$ $$+$$ $$^{2017}C_{1010}+…..+$$ $$^{2017}C_{2017}=2^{2017}$$
$$\Rightarrow \displaystyle \sum_{k=0}^{1008}$$ $$^{2017}C_{k}+$$ $$^{2017}C_{1008}+$$ $$^{2017}C_{1007}+…….+$$ $$^{2017}C_{0}=2^{2017}$$
$$[using\ ^{m}C_{r}=^{m}C_{m-r}]$$
$$\Rightarrow \displaystyle \sum_{k=0}^{1008}$$ $$^{2017}C_{k} $$ $$+ \displaystyle \sum_{1008}^{k=0}$$ $$^{2017}C_{k}= 2^{2017}$$
$$ \Rightarrow \displaystyle 2 \sum_{k=0}^{1008}$$ $$^{2017}C_{k}=2^{2017}$$
$$\Rightarrow \displaystyle \sum_{k=0}^{1008}$$ $$^{2017}C_{k}=2^{2016}$$
$$\Rightarrow\displaystyle \sum_{k=0}^{1008}$$ $$^{2017}C_{k}= (2^{1008})^{2}$$
$$\therefore$$ $$\lambda=2^{1008}$$
$$\lambda=2^{1008}$$
$$\lambda=2^{3}.2^{1005}$$
$$=2^{3} (2^{5})^{201}$$
$$=2^{3} (32)^{201}$$
$$=2^{3} (33-1)^{201}$$ [using binomial theorem ]
$$=8[^{201}C_{0} (33)^{201}+\, ^{201}C_{1} (33)^{200} (-1)+\, ^{201}C_{2} (33)^{199} (-1)^{2}+…..+\, ^{201}C_{201} (-1)^{201}]$$
$$=8[(33)^{201}+201 (33)^{200} (-1)+\, ^{201}C_{2} (33)^{199} (-1)^{2}+….+(-1)^{201}]$$
$$=[8 (33)^{201}+8.201 (300)^{200} (-1)+ 8\times ^{201}C_{2} (33)^{199} (-1)^{2}+….-8]$$
When $$\lambda$$ is divided by $$33$$ then the remainder is $$-8$$ which is negative we can also write the remainder as $$33-8=25$$
Question 6
1 / -0

Value of $$\sum _{ r=1 }^{ n }{ \left( \sum _{ m=0 }^{ r }{ { }_{ } } \right) } ^nC_r,^rC_m)$$ is equal to

A
$$2^n-1$$

B
$$3^n-1$$

C
$$3^n-2^n$$

D
$$None$$ $$of$$ $$these$$

Solution
Question 7
1 / -0

If $$(1+x)^n=C_0+C_1x+C_2x^2+....+C_nX^n$$, then the value of $$C^2_0+\dfrac{C^2_1}{2}+\dfrac{C^2_2}{3}+....+\dfrac{C^2_n}{n+1}$$ is equal to?

A
$$\dfrac{(2n-1)!}{\{(n+1)!\}^2}$$

B
$$\dfrac{(2n-1)!}{(n+1)!}$$

C
$$\dfrac{2n+1)!}{\{(n+1)!\}^2}$$

D
$$\dfrac{(2n)!}{(n+1)!}$$

Solution
Question 8
1 / -0

If $$(1+x)^{n}=C_{0}+C_{1}x+C_{2}x^{2}+...+C_{n}x^{n}$$ then the value of $$1^{2}C_{1}+2^{2}C_{2}+3^{2}C_{3}+...+n^{2}C_{n}$$ is

A
$$n(n+1)2^{n-2}$$

B
$$n(n+1)2^{n-1}$$

C
$$n(n+1)2^{n}$$

D
None of these

Solution
Question 9
1 / -0

The value of $${^{50}C_4}+\sum^6_{r=1}{^{56-r}C_3}$$ is?

A
$$^{55}C_4$$

B
$$^{55}C_3$$

C
$$^{56}C_3$$

D
$$^{56}C_4$$

Solution
Question 10
1 / -0

If n is odd natural number, then $$\sum _{ r=0 }^{ n }{ \cfrac { { (-1) }^{ r } }{ ^{ n }{ C }_{ r } } } $$ equals

A
0

B
1/n

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

D
None of these

Solution