Basic Concept of Permutations and Combinations-Test 1
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Subject :- QT
Chapter :- Basic Concept of Permutations and Combinations -Test 1
Questions :- 25
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Question 1 of 25
1 points
The number of triangles that can be formed by choosing the vertices from a set of 12 points ,seven of which lie on the same straight line is
Correct
$Thenumberoftrianglesthatcanbeformedfromasetof12points\phantom{\rule{0ex}{0ex}}=12{C}_{3}\phantom{\rule{0ex}{0ex}}\mathrm{sin}ce7pointsareonthesameline,thereforenotrianglecanbeformedfromthesepoints\phantom{\rule{0ex}{0ex}}i.e.numberoftriangles=12{C}_{3}\u20137{C}_{3}=220\u201335=185$
Incorrect
$Thenumberoftrianglesthatcanbeformedfromasetof12points\phantom{\rule{0ex}{0ex}}=12{C}_{3}\phantom{\rule{0ex}{0ex}}\mathrm{sin}ce7pointsareonthesameline,thereforenotrianglecanbeformedfromthesepoints\phantom{\rule{0ex}{0ex}}i.e.numberoftriangles=12{C}_{3}\u20137{C}_{3}=220\u201335=185$
Question 2 of 25
1 points
A code word is to consist of two English alphabets followed by 2 distinct numbers between 1 and 9.How many such code words are there?
Correct
The number of ways of filling first two places with english alphabets=26×25=650
The number of ways of filling last two places with distinct numbers=9×8=72
The number of code words that can be formed are=650×72=46800
Incorrect
The number of ways of filling first two places with english alphabets=26×25=650
The number of ways of filling last two places with distinct numbers=9×8=72
The number of code words that can be formed are=650×72=46800
Question 3 of 25
1 points
A boy has 3 library tickets and 8 books of his interest in the library of these 8,he does not want to borrow mathematics part-II unless mathematics part-I is also borrowed?In how many ways can he choose the three books to be borrowed?
Correct
There are two cases possible:
CASE 1:- When mathematics part-II is borrowed(i.e it means part-I has also been borrowed )
$numberofways=6{C}_{1}=6ways$
CASE 2:-when mathematics part-II is not borrowed(i.e.3 3 books are to be selected out of 7)
$numberofways=7{C}_{3}=35ways$
hence,total number of ways =35+6=41 ways
Incorrect
There are two cases possible:
CASE 1:- When mathematics part-II is borrowed(i.e it means part-I has also been borrowed )
$numberofways=6{C}_{1}=6ways$
CASE 2:-when mathematics part-II is not borrowed(i.e.3 3 books are to be selected out of 7)
$numberofways=7{C}_{3}=35ways$
hence,total number of ways =35+6=41 ways
Question 4 of 25
1 points
An exam paper consist of 12 questions divided into two parts A and B.Part A contains 7 questions and Part B contains 5 questions.A candidate is required to attempt 8 questions selecting at least 3 from each part.In how many ways can the candidate select the questions?
Correct
The candidate can select 8Q by selecting at least 3 from each part in the following ways :
$\left(a\right)3QfrompartAand5QfrompartB=7{C}_{3}\times 5{C}_{5}=35ways\phantom{\rule{0ex}{0ex}}\left(b\right)4QfrompartAandpartBeach=7{C}_{3}\times 5{C}_{4}=175ways\phantom{\rule{0ex}{0ex}}\left(c\right)5QfrompartAand3QfrompartB=7{C}_{5}\times 5{C}_{3}=210ways\phantom{\rule{0ex}{0ex}}$
hence,the total number of ways in which the candidate can select the question will be=35+175+210=420 ways
Incorrect
The candidate can select 8Q by selecting at least 3 from each part in the following ways :
$\left(a\right)3QfrompartAand5QfrompartB=7{C}_{3}\times 5{C}_{5}=35ways\phantom{\rule{0ex}{0ex}}\left(b\right)4QfrompartAandpartBeach=7{C}_{3}\times 5{C}_{4}=175ways\phantom{\rule{0ex}{0ex}}\left(c\right)5QfrompartAand3QfrompartB=7{C}_{5}\times 5{C}_{3}=210ways\phantom{\rule{0ex}{0ex}}$
hence,the total number of ways in which the candidate can select the question will be=35+175+210=420 ways
Question 5 of 25
1 points
A supreme court bench consist of 5 judges. in how many ways , the bench can give a majority division?
Correct
$majoritydecisioncanbetakenbyabenchof5judgesin\phantom{\rule{0ex}{0ex}}=5{C}_{3}+5{C}_{4}+5{C}_{5}=10+5+1=16ways$
Incorrect
$majoritydecisioncanbetakenbyabenchof5judgesin\phantom{\rule{0ex}{0ex}}=5{C}_{3}+5{C}_{4}+5{C}_{5}=10+5+1=16ways$
Question 6 of 25
1 points
Given:p(7,k)=60 ; p(7,k-3).then:
Correct
$p(7,k)=60p(7,k\u20133)\phantom{\rule{0ex}{0ex}}7{p}_{k}=60x7{p}_{k\u20133}\phantom{\rule{0ex}{0ex}}\frac{7!}{(7\u2013k)!}=60x\frac{7!}{[7\u2013(k\u20133\left)\right]!}\phantom{\rule{0ex}{0ex}}\frac{1}{(7\u2013k)!}=60x\frac{1}{[7\u2013(k\u20133)!}\phantom{\rule{0ex}{0ex}}\frac{(10\u2013k)!}{(7\u2013k)!}=60\phantom{\rule{0ex}{0ex}}\frac{(10\u2013k)(9\u2013k)(8\u2013k)(7\u2013k)!}{(7\u2013k)!}=60\phantom{\rule{0ex}{0ex}}{k}^{3}\u201327{k}^{2}+24k\u2013660=0\phantom{\rule{0ex}{0ex}}(k\u20135)({k}^{2}\u201322k+132)=0\phantom{\rule{0ex}{0ex}}k=5,\mathrm{sin}ce{k}^{2}\u201322k+132=0givesimaginaryroots$
Incorrect
$p(7,k)=60p(7,k\u20133)\phantom{\rule{0ex}{0ex}}7{p}_{k}=60x7{p}_{k\u20133}\phantom{\rule{0ex}{0ex}}\frac{7!}{(7\u2013k)!}=60x\frac{7!}{[7\u2013(k\u20133\left)\right]!}\phantom{\rule{0ex}{0ex}}\frac{1}{(7\u2013k)!}=60x\frac{1}{[7\u2013(k\u20133)!}\phantom{\rule{0ex}{0ex}}\frac{(10\u2013k)!}{(7\u2013k)!}=60\phantom{\rule{0ex}{0ex}}\frac{(10\u2013k)(9\u2013k)(8\u2013k)(7\u2013k)!}{(7\u2013k)!}=60\phantom{\rule{0ex}{0ex}}{k}^{3}\u201327{k}^{2}+24k\u2013660=0\phantom{\rule{0ex}{0ex}}(k\u20135)({k}^{2}\u201322k+132)=0\phantom{\rule{0ex}{0ex}}k=5,\mathrm{sin}ce{k}^{2}\u201322k+132=0givesimaginaryroots$
Question 7 of 25
1 points
The number of ways in which n books can be arranged on a shelf so that two particular books are not together is
Question 8 of 25
1 points
In how many ways can the letters of the word FAILURE be arranged so that the consonants may occupy only odd positions?
Correct
The word FAILURE have 7 letters out of which 3 letters F,L and R are consonants and 4 letters A,I,U and E are vowels.
The four positions to be filled up with consonants are indicated below:
1 2 3 4 5 6 7
F L R not filled
Since,there are only 3 consonants ,the total number of permutations are
$=4{P}_{3}=24$
Again,for the arrangement described above ,the four vowels can occupy the 4 remaining positions not occupied by consonants =4!=24 ways
hence the total number of arrangements are =24×24=576
Incorrect
The word FAILURE have 7 letters out of which 3 letters F,L and R are consonants and 4 letters A,I,U and E are vowels.
The four positions to be filled up with consonants are indicated below:
1 2 3 4 5 6 7
F L R not filled
Since,there are only 3 consonants ,the total number of permutations are
$=4{P}_{3}=24$
Again,for the arrangement described above ,the four vowels can occupy the 4 remaining positions not occupied by consonants =4!=24 ways
hence the total number of arrangements are =24×24=576
Question 9 of 25
1 points
5 bulbs of which 3 are defective are to be tried in two lights points in a dark room.In how many trials the room shall be lighted?
Correct
$Totalnumberoftrials=5{C}_{2}=10ways\phantom{\rule{0ex}{0ex}}no.oftrialsfornolightintheroom=3{C}_{2}=3\phantom{\rule{0ex}{0ex}}Theroomshallbelightedin=10\u20133=7ways$
Incorrect
$Totalnumberoftrials=5{C}_{2}=10ways\phantom{\rule{0ex}{0ex}}no.oftrialsfornolightintheroom=3{C}_{2}=3\phantom{\rule{0ex}{0ex}}Theroomshallbelightedin=10\u20133=7ways$
Question 10 of 25
1 points
In how many ways can a party of 4 men and 4 women be seated at a circular table ,so that no 2 women are adjacent?
Correct
The no.of ways in which 4 men can be seated at the circular table so that there is a vacant seat between every pair of men is=(4-1)!=3!=6 ways
hence,no.of ways in which 4 vacant seats can be occupied by 4 women =4!=24 ways
Required no of ways =6×24=144 ways
Incorrect
The no.of ways in which 4 men can be seated at the circular table so that there is a vacant seat between every pair of men is=(4-1)!=3!=6 ways
hence,no.of ways in which 4 vacant seats can be occupied by 4 women =4!=24 ways
Required no of ways =6×24=144 ways
Question 11 of 25
1 points
$Thevalueof\sum _{r=1}^{5}=5{C}_{r}is$
Correct
$\sum _{r=1}^{5}5{C}_{r}=5{C}_{1}+5{C}_{2}+5{C}_{3}+5{C}_{4}+5{C}_{5}\phantom{\rule{0ex}{0ex}}=5+10+10+5+1=31$
Incorrect
$\sum _{r=1}^{5}5{C}_{r}=5{C}_{1}+5{C}_{2}+5{C}_{3}+5{C}_{4}+5{C}_{5}\phantom{\rule{0ex}{0ex}}=5+10+10+5+1=31$
Question 12 of 25
1 points
$If6{P}_{r}=24x6{C}_{r},thenfindr$
Correct
$6{P}_{r}=24x6{C}_{r}\phantom{\rule{0ex}{0ex}}\frac{6!}{(6\u2013r)!}=24x\frac{6!}{r!x(6\u2013r)!}\phantom{\rule{0ex}{0ex}}4!=\frac{24}{r!}\phantom{\rule{0ex}{0ex}}r!=\frac{24}{4!}\phantom{\rule{0ex}{0ex}}r!=4!\phantom{\rule{0ex}{0ex}}r=4$
Incorrect
$6{P}_{r}=24x6{C}_{r}\phantom{\rule{0ex}{0ex}}\frac{6!}{(6\u2013r)!}=24x\frac{6!}{r!x(6\u2013r)!}\phantom{\rule{0ex}{0ex}}4!=\frac{24}{r!}\phantom{\rule{0ex}{0ex}}r!=\frac{24}{4!}\phantom{\rule{0ex}{0ex}}r!=4!\phantom{\rule{0ex}{0ex}}r=4$
Question 13 of 25
1 points
Find the number of combinations of the letters of the word COLLEGE taken four together
Question 14 of 25
1 points
How many words can be formed with the letters of the word ORIENTAL so that A and E always occupy odd places
Correct
$Thereare4oddplacesand2letters,hencethiscanbedonein=4{P}_{2}=12ways$
The remaining 6 letters i.e.O,R,I,N.T,L can be arranged in 6! ways
hence,the total no.of arrangements=6!x12=8640
Incorrect
$Thereare4oddplacesand2letters,hencethiscanbedonein=4{P}_{2}=12ways$
The remaining 6 letters i.e.O,R,I,N.T,L can be arranged in 6! ways
hence,the total no.of arrangements=6!x12=8640
Question 15 of 25
1 points
$If1000{C}_{98}=999{C}_{97}+X{C}_{901},findX$
Correct
$1000{C}_{98}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}\mathrm{sin}ce,n+1{C}_{r}=n{C}_{r}+n{C}_{r\u20131}\phantom{\rule{0ex}{0ex}}999{C}_{98}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}999{C}_{98}+999{C}_{97}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}999{C}_{98}+999{C}_{97}=999{C}_{97}+X{C}_{98}\phantom{\rule{0ex}{0ex}}X=999$
Incorrect
$1000{C}_{98}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}\mathrm{sin}ce,n+1{C}_{r}=n{C}_{r}+n{C}_{r\u20131}\phantom{\rule{0ex}{0ex}}999{C}_{98}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}999{C}_{98}+999{C}_{97}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}999{C}_{98}+999{C}_{97}=999{C}_{97}+X{C}_{98}\phantom{\rule{0ex}{0ex}}X=999$
Question 16 of 25
1 points
How many numbers greater than a million can be formed with the digits 4,5,5,0,4,5,3?
Correct
total no.of numbers that can be formed from the digits 4,5,5,0,4,5,3=7!/2!x3!=420 ways
out of these 420 numbers some will begin with 0 and are less than one million ,so they are to be rejected
number of numbers beginning with
0=6!/2!x3!=60 ways
hence required number of numbers=420-60=360 ways
Incorrect
total no.of numbers that can be formed from the digits 4,5,5,0,4,5,3=7!/2!x3!=420 ways
out of these 420 numbers some will begin with 0 and are less than one million ,so they are to be rejected
number of numbers beginning with
0=6!/2!x3!=60 ways
hence required number of numbers=420-60=360 ways
Question 17 of 25
1 points
A building contractor needs 3 helpers and 10 men apply.In how many ways can these selections take place?
Correct
since there is no regard for order,the contractor can select any of the 3 helpers out of 10 men.
this can be done in
$10{C}_{3}ways=120ways$
Incorrect
since there is no regard for order,the contractor can select any of the 3 helpers out of 10 men.
this can be done in
$10{C}_{3}ways=120ways$
Question 18 of 25
1 points
There are 3 blue balls,4 red balls and 5 green balls.in how many ways can they be arranged in a row ?
Correct
$No.ofway\mathrm{sin}whichtheseballscanbearranged\phantom{\rule{0ex}{0ex}}=\frac{12!}{3!x4!x5!}\phantom{\rule{0ex}{0ex}}=27720ways$
Incorrect
$No.ofway\mathrm{sin}whichtheseballscanbearranged\phantom{\rule{0ex}{0ex}}=\frac{12!}{3!x4!x5!}\phantom{\rule{0ex}{0ex}}=27720ways$
Question 19 of 25
1 points
$IfC(n,r):C(n,r+1)=1:2andC(n,r+1):C(n,r+2)=2:3,determinethevalueofnandr$
Question 20 of 25
1 points
6 seats if articled clerks are vacant in a Chartered Accountant Firm.how many different batches of candidates can be chosen out of 10 candidates?
Correct
The no.of ways in which 6 articles clerks can be selected out of 10 candidates
$10{C}_{6}=210ways$
Incorrect
The no.of ways in which 6 articles clerks can be selected out of 10 candidates
$10{C}_{6}=210ways$
Question 21 of 25
1 points
6 persons A,B,C,D,E and F are to be seated at a circular table.in how many ways can this be done ,if A must always have either B or C on his right and B must always have either C or D on his right ?
Correct
Using the given restrictions ,we must have AB or AC and BC or BD.
therefore,we have the following alternatives:
(i)ABC,D,E,F which gives (4-1)! or 31 ways
(ii)ABD,C,e,f which gives (4-1)! or 31 ways
(iii)AC,BD,E,F which gives (4-1)! or 31 ways
hence,the total number of ways are =3!+3!+3!=6+6+6=18 ways
Incorrect
Using the given restrictions ,we must have AB or AC and BC or BD.
therefore,we have the following alternatives:
(i)ABC,D,E,F which gives (4-1)! or 31 ways
(ii)ABD,C,e,f which gives (4-1)! or 31 ways
(iii)AC,BD,E,F which gives (4-1)! or 31 ways
hence,the total number of ways are =3!+3!+3!=6+6+6=18 ways
Question 22 of 25
1 points
$Ifn{p}_{r}=n{p}_{r+1}andn{c}_{r}=n{c}_{r\u20131}thenfindthevalueofn$
Correct
$n{p}_{r}=n{p}_{r+1}\Rightarrow n=r+1...........\left(1\right)\phantom{\rule{0ex}{0ex}}n{c}_{r}=n{c}_{r\u20131}\Rightarrow n=2r\u20131..........\left(2\right)\phantom{\rule{0ex}{0ex}}from12weget\phantom{\rule{0ex}{0ex}}r+1=2r\u20131\phantom{\rule{0ex}{0ex}}r=2\phantom{\rule{0ex}{0ex}}hence,n=2+1=3$
Incorrect
$n{p}_{r}=n{p}_{r+1}\Rightarrow n=r+1...........\left(1\right)\phantom{\rule{0ex}{0ex}}n{c}_{r}=n{c}_{r\u20131}\Rightarrow n=2r\u20131..........\left(2\right)\phantom{\rule{0ex}{0ex}}from12weget\phantom{\rule{0ex}{0ex}}r+1=2r\u20131\phantom{\rule{0ex}{0ex}}r=2\phantom{\rule{0ex}{0ex}}hence,n=2+1=3$
Question 23 of 25
1 points
In how many ways a committee of 6 members cab be formed from a group of 7 boys and 4 girls having at least 2 girls in the committee.
Question 24 of 25
1 points
Number of ways of painting a face of a cube by 6 colours is_____
Correct
number of ways of painting a face of a cube by 6 colours is 6,since any of the 6 colours can be used to paint the face of the cube.
Incorrect
number of ways of painting a face of a cube by 6 colours is 6,since any of the 6 colours can be used to paint the face of the cube.
Question 25 of 25
1 points
$if\_\_\_\_\_18{C}_{r}=18{C}_{r+2}findthevalueofr{c}_{5}$
I