Permutations and Combinations Test 38

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Permutations and Combinations Test 38

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Weekly Quiz Competition

Question 1
1 / -0

$$^{60}C_{60}=?$$

A
$$60!$$

B
$$1$$

C
$$\dfrac{1}{60}$$

D
None of these

Solution

$$^{60}C_{60}=1$$ $$[\because {^nC_n}=1]$$.
Question 2
1 / -0

If $$^{n}C_{18}={^nC_{12}}$$, then $$^{32}C_n=?$$

A
$$248$$

B
$$496$$

C
$$992$$

D
None of these

Solution

$$^nC_p={^nC_q}\Rightarrow p+q=n$$.
$$^nC_{18}={^{n}C_{12}}\Rightarrow n=(18+12)=30$$.
$$\therefore {^{32}C_n}={^{32}C_{30}}={^{32}C_2}=\dfrac{32\times 31}{2}=496$$.
Question 3
1 / -0

There are $$10$$ points in a plane, out of which $$4$$ points are collinear. The number of triangles formed with vertices as these point is?

A
$$20$$

B
$$120$$

C
$$116$$

D
None of these

Solution

Number of triangles obtained from $$10$$ points $$={^{10}C_3}=\dfrac{10\times 9\times 8}{3\times 2\times 1}=120$$.
Number of triangles obtained from $$4$$ points $$={^4C_3}={^4C_1}=4$$
But, these $$4$$ points being collinear will give no triangle.
$$\therefore$$ required number of triangles $$=(120-4)=116$$.
Question 4
1 / -0

When simplified, the expression
$$^{ 47 }{ C }_{ 4 }+\sum _{ j=1 }^{ 5 }$$ $$^{ 52-j }{ C }_{ 3 } $$ equals to

A
$$^{ 47 }{ C }_{ 5 }$$

B
$$^{ 49 }{ C }_{ 4 }$$

C
$$^{ 52 }{ C }_{ 5 }$$

D
$$^{ 52 }{ C }_{ 4 }$$

Solution
Question 5
1 / -0

How many $$3$$-digit numbers are there?

A
$$648$$

B
$$729$$

C
$$900$$

D
$$1000$$

Solution

The hundreds place can be filled by any of the $$9$$ non zero digits.
So, there are $$9$$ ways of filling this place. The tens place can be filled by any of the $$10$$ digits. So, there are $$10$$ ways of filling it.
The units place can be filled by any of the $$10$$ digits. So, there are $$10$$ ways of filling it.
$$\therefore$$ total number of $$3$$-digit numbers$$=(9\times 10\times 10)=900$$.
Question 6
1 / -0

If $$^{ n }{ C }_{ r-1 }=10,$$ $$^{ n }{ C }_{ r }=45,$$ and $$^{ n }{ C }_{ r+1 }=120,$$ then r equals

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 7
1 / -0

In an examination, a candidate has to pass in each of the five subjects. In how many ways can he fail?

A
$$5$$

B
$$10$$

C
$$21$$

D
$$31$$

Solution

The candidate can fail by failing in $$1$$ or $$2$$ or $$3$$ or $$4$$ or $$5$$ subjects out of $$5$$ in each case.
$$\therefore$$ required number of ways$$={^5C_1}+{^5C_2}+{^5C_3}+{^5C_4}+{^5C_5}$$
$$={^5C_1}+{^5C_2}+{^5C_{(5-3)}}+{^5C_{(5-4)}}+1$$
$$={^5C_1}+{^5C_2}+{^5C_2}+{^5C_1}+1$$
$$=2({^5C_1}+{^5C_2})+1=2\left(5+\dfrac{5\times 4}{2\times 1}\right)+1=(30+1)=31$$.
Question 8
1 / -0

A committee of $$5$$ is to be formed out of $$6$$ gents and $$4$$ ladies. In how many ways can this be done when each committee may have at the most $$2$$ ladies?

A
$$120$$

B
$$160$$

C
$$180$$

D
$$186$$

Solution

We may have:
(i) ($$1$$ lad out of $$4$$) and ($$4$$ gents out of $$6$$)
or (ii) ($$2$$ ladies out of $$4$$) and ($$3$$ gents out of $$6$$).
$$\therefore$$ required number of ways
$$=(^4C_1\times {^6C_4})+(^4C_2\times {^6C_3})\\=(4\times {^6C_2})+\left(\dfrac{4\times 3}{2\times 1}\times \dfrac{6\times 5\times 4}{3\times 2\times 1}\right)$$
$$=\left(4\times \dfrac{6\times 5}{2\times 1}\right)+(6\times 20)\\=(60+120)=180$$.
Question 9
1 / -0

Out of $$7$$ consonants and $$4$$ vowels, how many words of $$3$$ consonants and $$2$$ vowels can be formed?

A
$$330$$

B
$$1050$$

C
$$6300$$

D
$$25200$$

Solution

Number of ways of selecting $$3$$ consonants out of $$7$$ and $$2$$ vowels out of $$4$$ $$=(^7C_3\times {^4C_2})=\left(\dfrac{7\times 6\times 5}{3\times 2\times 1}\times \dfrac{4\times 3}{2\times 1}\right)=210$$.
Now, $$5$$ letters can be arranged among themselves in $$5!$$ ways $$=120$$ ways.
Required number of words$$=(210\times 120)=25200$$.
Question 10
1 / -0

In how many ways can a committee of $$5$$ members be selected from $$6$$ men and $$5$$ ladies, consisting of $$3$$ men and $$2$$ ladies?

A
$$25$$

B
$$50$$

C
$$100$$

D
$$200$$

Solution

Number of ways of selecting $$3$$ men out of $$6$$ and $$2$$ ladies out of $$5$$
$$=(^6C_3\times {^5C_2})=\left(\dfrac{6\times 5\times 4}{3\times 2\times 1}\times \dfrac{5\times 4}{2\times 1}\right)=200$$.

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 38

Result

Permutations and Combinations Test 38

Score

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Rank

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SHARING IS CARING

Question 1

$$60!$$

$$1$$

$$\dfrac{1}{60}$$

None of these

Solution

Question 2

$$248$$

$$496$$

$$992$$

None of these

Solution

Question 3

$$20$$

$$120$$

$$116$$

None of these

Solution

Question 4

$$^{ 47 }{ C }_{ 5 }$$

$$^{ 49 }{ C }_{ 4 }$$

$$^{ 52 }{ C }_{ 5 }$$

$$^{ 52 }{ C }_{ 4 }$$

Solution

Question 5

$$648$$

$$729$$

$$900$$

$$1000$$

Solution

Question 6

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 7

$$5$$

$$10$$

$$21$$

$$31$$

Solution

Question 8

$$120$$

$$160$$

$$180$$

$$186$$

Solution

Question 9

$$330$$

$$1050$$

$$6300$$

$$25200$$

Solution

Question 10

$$25$$

$$50$$

$$100$$

$$200$$

Solution

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