Permutations and Combinations Test 37

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Permutations and Combinations Test 37

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Weekly Quiz Competition

Question 1
1 / -0

There are $$31$$ objects in a bag in which $$10$$ are identical, then the number of ways of choosing $$10$$ objects from bag is?

A
$$2^{20}$$

B
$$2^{20}-1$$

C
$$2^{20}+1$$

D
$$2^{21}$$

Solution

$$^{21}C_0+{^{21}C_1}+{^{21}C_2}+.....+{^{21}C_{10}}=\dfrac{2^{21}}{2}=2^{20}$$.
Question 2
1 / -0

The coefficient of $$x^{18}$$ in the expansion of $$(1+x)(1-x)^{10}\{(1+x+x^2)^9\}$$ is?

A
$$84$$

B
$$126$$

C
$$-42$$

D
$$42$$

Solution

Coefficient of $$x^{18}$$ in $$(1+x)(1-x)^{10}(1+x+x^2)^9$$
$$=$$Coefficient of $$x^{18}$$ in $$(1-x)^2\{(1-x)(1+x+x^2)\}^9$$
$$=$$Coefficient of $$x^{18}$$ in $$(1-x^2)(1-x^3)^9$$
$$={^9C_6}=0=84$$.
Question 3
1 / -0

A team of three persons with at least one boy and atleast one girl is to be formed from $$5$$ boys and $$n$$ girls. If the number of sum teams is $$1750$$, then the value of $$n$$ is

A
$$24$$

B
$$28$$

C
$$27$$

D
$$25$$

Solution

Given $$5$$ boys, $$n$$ girls
$$(1B,2G)+(2B,1G)$$
$${ _{ }^{ 5 }{ C } }_{ 1 }.{ _{ }^{ n }{ C } }_{ 2 }+{ _{ }^{ 5 }{ C } }_{ 2 }.{ _{ }^{ n }{ C } }_{ 1 }=1750\Rightarrow 5.\cfrac { n(n-1) }{ 2 } +10.n=1750\Rightarrow \cfrac { n(n-1) }{ 2 } +2n=350\Rightarrow { n }^{ 2 }-n+4n=700$$
$${ n }^{ 2 }+3n-700=0\Rightarrow (n+28)(n-25)=0\Rightarrow n=25,-28$$
n $$\neq -28$$ as number of teams cannot be negative
Question 4
1 / -0

If $$\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{x}{8!}$$, then $$x=?$$

A
$$32$$

B
$$48$$

C
$$56$$

D
$$64$$

Solution

$$\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{x}{8!}\Rightarrow \dfrac{8\times 7}{8\times 7\times (6!)}+\dfrac{8}{8\times (7!)}=\dfrac{x}{8!}$$
$$\Rightarrow \dfrac{56}{8!}+\dfrac{8}{8!}=\dfrac{x}{8!}\Rightarrow x=56+8=64$$.
Question 5
1 / -0

$$^{36}C_{34}=?$$

A
$$1224$$

B
$$612$$

C
$$630$$

D
None of these

Solution

$$^{36}C_{34}={^{36}C_{(36-34)}}={^{36}C_2}=\dfrac{36\times 35}{2}=630$$.
Question 6
1 / -0

$$\dfrac{^nC_r}{^nC_{r-1}}=?$$

A
$$\dfrac{n-r}{r}$$

B
$$\dfrac{n-r-1}{r}$$

C
$$\dfrac{n-r+1}{r}$$

D
None of these

Solution

$$\dfrac{^nC_r}{^nC_{r-1}}\\=\dfrac{n!}{(r!)\times (n-r)!}\times \dfrac{(r-1)!\times (n-r+1)!}{n!}\\$$
$$=\dfrac{(r-1)!\times (n-r+1)\times (n-r)!}{r\cdot (r-1)!\times (n-r)!}\\=\dfrac{(n-r+1)}{r}$$.
Question 7
1 / -0

There are $$10$$ points in a plane, out of which $$4$$ points are collinear. The number of line segments obtained from the pairs of these points is?

A
$$39$$

B
$$40$$

C
$$41$$

D
$$45$$

Solution

Number of line segments formed by joining pairs of points out of $$10$$ $$={^{10}C_2}=\dfrac{10\times 9}{2}=45$$.
Number of line segments formed by joining pairs of $$4$$ points $$={^4C_2}=\dfrac{4\times 3}{2}=6$$.
But, these points being collinear give only one line.
$$\therefore$$ required number of line segments$$=(45-6+1)=40$$.
Question 8
1 / -0

If $$^{n}C_3=220$$, then $$n=?$$

A
$$9$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

$$^nC_3=220\Rightarrow \dfrac{n(n-1)(n-2)}{6}=220$$
$$\Rightarrow n(n-1)(n-2)=1320$$
$$\Rightarrow n=12$$ $$[\because 12\times 11\times 10=1320]$$.
Question 9
1 / -0

If $$^nC_{10}={^{n}C_{14}}$$, then $$n=?$$

A
$$4$$

B
$$24$$

C
$$14$$

D
$$10$$

Solution

$$^nC_p={^nC_q}\Rightarrow p+q=n$$.
$$\therefore {^nC_{10}}={^{n}C_{14}}\Rightarrow n=(10+14)=24$$.
Question 10
1 / -0

If $$^{n}C_r+{^nC_{r+1}}={^{n+1}C_x}$$, then $$x=?$$

A
$$r-1$$

B
$$r$$

C
$$r+1$$

D
$$n$$

Solution

We know that $$^{n}C_r+{^nC_{r+1}}={^{n+1}C_{r+1}}$$.
So, $$x=(r+1)$$.

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 37

Result

Permutations and Combinations Test 37

Score

-

Rank

-

SHARING IS CARING

Question 1

$$2^{20}$$

$$2^{20}-1$$

$$2^{20}+1$$

$$2^{21}$$

Solution

Question 2

$$84$$

$$126$$

$$-42$$

$$42$$

Solution

Question 3

$$24$$

$$28$$

$$27$$

$$25$$

Solution

Question 4

$$32$$

$$48$$

$$56$$

$$64$$

Solution

Question 5

$$1224$$

$$612$$

$$630$$

None of these

Solution

Question 6

$$\dfrac{n-r}{r}$$

$$\dfrac{n-r-1}{r}$$

$$\dfrac{n-r+1}{r}$$

None of these

Solution

Question 7

$$39$$

$$40$$

$$41$$

$$45$$

Solution

Question 8

$$9$$

$$10$$

$$11$$

$$12$$

Solution

Question 9

$$4$$

$$24$$

$$14$$

$$10$$

Solution

Question 10

$$r-1$$

$$r$$

$$r+1$$

$$n$$

Solution

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