Permutations and Combinations Test 46

Result

Permutations and Combinations Test 46

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$12$$ people came for a carroms tournament. All were divided into pairs of $$2$$ each. After all the matches if it found that half of the selected pairs had to play three matches to decide the winner in a best of three process, how many matches have been held?

A
$$165$$

B
$$185$$

C
$$198$$

D
$$132$$

Solution
Question 2
1 / -0

If $$^{19}C_{3r} = ^{19}C_{r+3} $$, then r is equal to:

A
$$5$$

B
$$4$$

C
$$3$$

D
$$2$$

Solution

$$^{ 19 }{ C }_{ 3r }=^{ 19 }{ C }_{ r+3 }$$
We Know
$$^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }\\ \Rightarrow r+(n-r)=n\\ \Rightarrow 3r+r+3=19\\ \Rightarrow 4r=16\\ \Rightarrow r=4 $$
Therefore The value of $$r$$ is $$4$$
Question 3
1 / -0

How many alphabets need to be there in a language if one were to make $$1$$ million distinct $$3$$ digit initials using the alphabets of the language?

A
$$10$$

B
$$100$$

C
$$56$$

D
$$26$$

Solution

Digits with initials$$=3$$
Total number of initials$$=1million=1000000$$
Let the alphabets in language be $$x$$
initials
$$={ x }^{ 3 }=1000000={ 10 }^{ 6 }\\ \Rightarrow x={ 10 }^{ 2 }$$
Therefore $$100$$ alphabets are present
Question 4
1 / -0

After every get-together every person present shakes the hand of every other person. If there were 105 handshakes in all, how many persons were present in the party?

A
$$16$$

B
$$15$$

C
$$13$$

D
$$14$$

Solution

Total number of handshakes $$=105$$
Let number of person be $$=x$$
Number of handshakes
$$=\cfrac { x(x-1) }{ 2 } =105\\ \Rightarrow x(x-1)=210\\ x=15$$
Hence the correct answer is $$15$$
Question 5
1 / -0

$$8064$$ is resolved into all possible product of two factors. Find the number of ways in which this can be done?

A
$$24$$

B
$$21$$

C
$$20$$

D
None of these

Solution

$$8064=2^{ 7 }\times { 3 }^{ 2 }\times 7$$
Resolution of a number into factors (and thier products).
depends on powers of primes.
Number of possible products are
$$=\left( 7+1 \right) \left( 2+1 \right) \left( 1 \right) \\ =8\times 3\times 1\\ =24$$
Question 6
1 / -0

How many straight lines can be formed from $$11$$ points in a plane out of which no three points are collinear?

A
$$66$$

B
$$50$$

C
$$55$$

D
$$60$$

Solution

Number of points $$=11$$
No three are collinear
For a straight lines Two end points are needed
Number of straight lines needed
$$\Rightarrow ^{ 11 }{ C }_{ 2 }\\ \Rightarrow \cfrac { 11! }{ 2!\times 9! } \\ \Rightarrow \cfrac { 11\times 10 }{ 2 } \\ \Rightarrow 55 lines$$
Option $$: C$$
Question 7
1 / -0

A college offers $$7$$ courses in the morning and $$5$$ in the evening. Find possible number of choices with the student who want to study one course in the morning and one in the evening.

A
$$35$$

B
$$12$$

C
$$49$$

D
$$25$$

Solution

Number of courses in the morning $$=7$$
Selecting one of the $$7=^{ 7 }{ C }_{ 1 }$$ Ways $$=7$$ ways
Number of courses in the evening $$= 5$$
Selecting one of the $$5=^{ 5 }{ C }_{ 1 } $$ Ways$$ = 5$$ ways
Total number of ways$$ =7\times 5$$ Ways $$=35$$ Ways
Question 8
1 / -0

When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?

A
$$25$$

B
$$41$$

C
$$22$$

D
$$42$$

Solution

Total number of coins $$=6$$

(1) '0' heads $$=1$$ chance$$ \left( ^{ 6 }{ C }_{ 0 } \right) $$

(2) '1' heads $$=6$$ chance$$ \left( ^{ 6 }{ C }_{ 1 } \right) $$

(3) '2' heads$$ =15$$ chance$$ \left( ^{ 6 }{ C }_{ 2 } \right) $$

(4) '3' heads$$ =20$$ chance $$ \left( ^{ 6 }{ C }_{ 3 } \right) $$

Hence total outcome $$=42$$
Question 9
1 / -0

The number of ways in which four letters can be selected from the word 'DEGREE' is

A
$$7$$

B
$$6$$

C
$$\dfrac{6!}{3!}$$

D
None of these

Solution

There are $$6$$ letters in the word 'DEGREE', namely $$3\text{E's}$$ and $$\text{D, G, R}$$. Four letters out of six can be selected in the following ways
(i) $$3$$ like letters and $$1$$ different, viz. $$\text{EEE + D, G, R}$$
(ii) $$2$$ like letters and $$2$$ different, viz. $$\text{EEE}\ +$$ any two of $$\text{D, G, R}$$.
(iii) All different letters, viz. $$\text{EDGR}$$.
So, the total number of ways
$$=^3C_1 + ^3C_2 + 1 = 3 + 3 + 1 = 7$$
Question 10
1 / -0

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

A
$$3$$

B
$$2$$

C
$$0$$

D
$$1$$

Solution

$$^{ 15 }{ C }_{ 3r }=^{ 15 }{ C }_{ r+3 }$$
We Know
$$^{ 15 }{ C }_{ 3r }=^{ 15 }{ C }_{ r+3 }\\ ^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }\\ n=r+(n-r)\\ =3r+r+3\\ =4r+3\\ \Rightarrow 4r+3=15\\ \Rightarrow 4r=12\\ \Rightarrow r=3$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 46

Result

Permutations and Combinations Test 46

Score

-

Rank

-

SHARING IS CARING

Question 1

$$165$$

$$185$$

$$198$$

$$132$$

Solution

Question 2

$$5$$

$$4$$

$$3$$

$$2$$

Solution

Question 3

$$10$$

$$100$$

$$56$$

$$26$$

Solution

Question 4

$$16$$

$$15$$

$$13$$

$$14$$

Solution

Question 5

$$24$$

$$21$$

$$20$$

None of these

Solution

Question 6

$$66$$

$$50$$

$$55$$

$$60$$

Solution

Question 7

$$35$$

$$12$$

$$49$$

$$25$$

Solution

Question 8

$$25$$

$$41$$

$$22$$

$$42$$

Solution

Question 9

$$7$$

$$6$$

$$\dfrac{6!}{3!}$$

None of these

Solution

Question 10

$$3$$

$$2$$

$$0$$

$$1$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.