Basic Concept of Permutations and Combinations-Test 2
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Subject :- QT
Chapter :- Basic Concept of Permutations and Combinations -Test 2
Questions :- 25
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Question 1 of 25
1 points
The number of words which can be formed by letters of the word ALLAHABAD is
Correct
given word ALLAHABAD
no.of ways =n!/p!q!
total no.of letter n=9
no of A p=4
no of L q=2
no of words =9!/4!x2!=9x8x7x6x5x4!/4!x2!=7560
Incorrect
given word ALLAHABAD
no.of ways =n!/p!q!
total no.of letter n=9
no of A p=4
no of L q=2
no of words =9!/4!x2!=9x8x7x6x5x4!/4!x2!=7560
Question 2 of 25
1 points
The number of parallelograms,formed from a set of 6 parallel lines intersecting another set of 4 parallel lines is
Correct
No.of parallelogram are formed from a set of m parallel lines intersecting another set of n
$parallellines=m{c}_{2}xn{c}_{2}\phantom{\rule{0ex}{0ex}}here,m=6,n=4\phantom{\rule{0ex}{0ex}}=6{c}_{2}x4{c}_{2}\phantom{\rule{0ex}{0ex}}=6x5/2x1x4x3/2x1\phantom{\rule{0ex}{0ex}}=15x6=90$
Incorrect
No.of parallelogram are formed from a set of m parallel lines intersecting another set of n
$parallellines=m{c}_{2}xn{c}_{2}\phantom{\rule{0ex}{0ex}}here,m=6,n=4\phantom{\rule{0ex}{0ex}}=6{c}_{2}x4{c}_{2}\phantom{\rule{0ex}{0ex}}=6x5/2x1x4x3/2x1\phantom{\rule{0ex}{0ex}}=15x6=90$
Question 3 of 25
1 points
$If10{C}_{3}+2.10{C}_{4}+10{C}_{5}=n{C}_{5},thenvalueofnis$
Correct
$If10{C}_{3}+2.10{C}_{4}+10{C}_{5}=n{C}_{5}\phantom{\rule{0ex}{0ex}}(10{C}_{3}+10{C}_{4})+(10{C}_{4}+10{C}_{5})=n{C}_{5}\phantom{\rule{0ex}{0ex}}10+1{C}_{5}+10+1{C}_{4}=n{C}_{5}\phantom{\rule{0ex}{0ex}}11{C}_{5}+11{C}_{4}=n{C}_{5}\phantom{\rule{0ex}{0ex}}11+1{C}_{5}=n{C}_{5}\phantom{\rule{0ex}{0ex}}12{C}_{5}=n{C}_{5}\phantom{\rule{0ex}{0ex}}n=12$
Incorrect
$If10{C}_{3}+2.10{C}_{4}+10{C}_{5}=n{C}_{5}\phantom{\rule{0ex}{0ex}}(10{C}_{3}+10{C}_{4})+(10{C}_{4}+10{C}_{5})=n{C}_{5}\phantom{\rule{0ex}{0ex}}10+1{C}_{5}+10+1{C}_{4}=n{C}_{5}\phantom{\rule{0ex}{0ex}}11{C}_{5}+11{C}_{4}=n{C}_{5}\phantom{\rule{0ex}{0ex}}11+1{C}_{5}=n{C}_{5}\phantom{\rule{0ex}{0ex}}12{C}_{5}=n{C}_{5}\phantom{\rule{0ex}{0ex}}n=12$
Question 4 of 25
1 points
The number of numbers between 1000 and 10000 which can be formed by the digits 1,2,3,4,5,6 without repetition is
Correct
1 TH H T U
2 6x 5 x4 x3 =360
3
4
5
6
no.of numbers between 1000 and 10000 using the digit 1,2,3,4,5,6=360
Incorrect
1 TH H T U
2 6x 5 x4 x3 =360
3
4
5
6
no.of numbers between 1000 and 10000 using the digit 1,2,3,4,5,6=360
Question 5 of 25
1 points
The maximum number of points of inter section of 10 circles will be
Correct
$Themaximumno.ofpointsofintersectionofncircle\phantom{\rule{0ex}{0ex}}=n{p}_{2}\phantom{\rule{0ex}{0ex}}=10{p}_{2}\phantom{\rule{0ex}{0ex}}=10x9\phantom{\rule{0ex}{0ex}}=90$
Incorrect
$Themaximumno.ofpointsofintersectionofncircle\phantom{\rule{0ex}{0ex}}=n{p}_{2}\phantom{\rule{0ex}{0ex}}=10{p}_{2}\phantom{\rule{0ex}{0ex}}=10x9\phantom{\rule{0ex}{0ex}}=90$
Question 6 of 25
1 points
7 books are to be arranged in such a way so that two particular books are always at first and last place.Final the number of arrangements.
Correct
Since 2 particular books are to be kept always at the first and last place ,so if we fix places,the remaining 5 books can be arranged in 5! ways
those,2 books can also change their places in 2! ways
the total number of arrangements are=5!x2!=120×2=240 ways
Incorrect
Since 2 particular books are to be kept always at the first and last place ,so if we fix places,the remaining 5 books can be arranged in 5! ways
those,2 books can also change their places in 2! ways
the total number of arrangements are=5!x2!=120×2=240 ways
Question 7 of 25
1 points
Find the number of arrangements in which the letter of the word MONDAY be arranged so that the words thus formed begin with M and do not end with N
Correct
In the first place only M can come.therefore,no.of ways of filling first place is 1!=1 way
In the last place ,any letter other than M and N can come. therefore ,no of ways in which last place can be filled is 4 ways i.e 6-2=4 ways
The remaining 4 places have no restrictions.They can be filled by any of the remaining 4 letters.so,the no.of ways of filling those 4 place are 4! ways
so,the total no.of words that can be formed are=1x4x4! ways =4×24=96 ways
Incorrect
In the first place only M can come.therefore,no.of ways of filling first place is 1!=1 way
In the last place ,any letter other than M and N can come. therefore ,no of ways in which last place can be filled is 4 ways i.e 6-2=4 ways
The remaining 4 places have no restrictions.They can be filled by any of the remaining 4 letters.so,the no.of ways of filling those 4 place are 4! ways
so,the total no.of words that can be formed are=1x4x4! ways =4×24=96 ways
Question 8 of 25
1 points
In how many ways can 17 billiard balls be arranged if 7 of them are black,6 red and 4 white?
Correct
$Numberofwaysofarranging17{P}_{17}i.e.17!ways\phantom{\rule{0ex}{0ex}}but7ofthemareblack,6redand4white\phantom{\rule{0ex}{0ex}}no.ofwaysofarrengingtheballs=\frac{17!}{7!x6!x4!}=4084080$
Incorrect
$Numberofwaysofarranging17{P}_{17}i.e.17!ways\phantom{\rule{0ex}{0ex}}but7ofthemareblack,6redand4white\phantom{\rule{0ex}{0ex}}no.ofwaysofarrengingtheballs=\frac{17!}{7!x6!x4!}=4084080$
Question 9 of 25
1 points
Out of 4 gents and 6 ladies a committee is to be formed find the number of ways the committee can be formed such that it comprises of at least 2 gents and at least the number of ladies should be double of gents.
Correct
$no.ofwaysofselecting2gents=4{C}_{2}\phantom{\rule{0ex}{0ex}}no.ofwaysofselectingladiessothatatleastthenumberofladiesis\phantom{\rule{0ex}{0ex}}doubleofgents=6C4+6{C}_{5}+6{C}_{6}\phantom{\rule{0ex}{0ex}}totalways=4{C}_{2}x(6{C}_{4}+6{C}_{5}+6{C}_{6})\phantom{\rule{0ex}{0ex}}=6x(15+6+1)\phantom{\rule{0ex}{0ex}}=132ways$
Incorrect
$no.ofwaysofselecting2gents=4{C}_{2}\phantom{\rule{0ex}{0ex}}no.ofwaysofselectingladiessothatatleastthenumberofladiesis\phantom{\rule{0ex}{0ex}}doubleofgents=6C4+6{C}_{5}+6{C}_{6}\phantom{\rule{0ex}{0ex}}totalways=4{C}_{2}x(6{C}_{4}+6{C}_{5}+6{C}_{6})\phantom{\rule{0ex}{0ex}}=6x(15+6+1)\phantom{\rule{0ex}{0ex}}=132ways$
Question 10 of 25
1 points
In a bag there were 5 white,3 red and 2 black balls.3 balls are drawn at a time what is the probability that the 3 balls drawn are white?
Correct
no.of ways of drawing 3 balls at a time=120 ways
no.of ways of drawing 3 white balls out of 5 white balls =10 ways
total no.of ways=favourable cases/total no.of cases=10/129=1/12
Incorrect
no.of ways of drawing 3 balls at a time=120 ways
no.of ways of drawing 3 white balls out of 5 white balls =10 ways
total no.of ways=favourable cases/total no.of cases=10/129=1/12
Question 11 of 25
1 points
6 points are on a circle.The number of quadrilaterals that can be formed are:
Correct
$no.ofquadrilateralsthatcanbeformedfrom6pointsonacircle\phantom{\rule{0ex}{0ex}}=6{C}_{4}\phantom{\rule{0ex}{0ex}}=\frac{6!}{2!4!}\phantom{\rule{0ex}{0ex}}=30/2\phantom{\rule{0ex}{0ex}}=15$
Incorrect
$no.ofquadrilateralsthatcanbeformedfrom6pointsonacircle\phantom{\rule{0ex}{0ex}}=6{C}_{4}\phantom{\rule{0ex}{0ex}}=\frac{6!}{2!4!}\phantom{\rule{0ex}{0ex}}=30/2\phantom{\rule{0ex}{0ex}}=15$
Question 12 of 25
1 points
The number of ways of arranging 6 boys and 4 girls in a row so that all 4 girls are together is
Correct
let us assume 4 girls to be seated together so they can be arranged in 4! ways
now,if we assume 4 girls to be one single group
total number of ways of arranging boys and girls is 7!
required numbers of ways are 7!x4!
Incorrect
let us assume 4 girls to be seated together so they can be arranged in 4! ways
now,if we assume 4 girls to be one single group
total number of ways of arranging boys and girls is 7!
required numbers of ways are 7!x4!
Question 13 of 25
1 points
How many members not exceeding 1000 can be made from the digits 1,2,3,4,5,6,7,8,9 if repetition is not allowed
Correct
total no.of 2 digits that can be formed=9×8=72
total no.of 3 digits that can be formed=9x8x7=504
total no.of 1 digits that can be formed=9
total numbers that can be formed=9+72+504=585
Incorrect
total no.of 2 digits that can be formed=9×8=72
total no.of 3 digits that can be formed=9x8x7=504
total no.of 1 digits that can be formed=9
total numbers that can be formed=9+72+504=585
Question 14 of 25
1 points
A garden having 6 tall trees in a row.in how many ways 5 children stand ,one in a gap between the tree in order to pose for a photograph.
Correct
5 places between the trees
0 0 0 0 0
X X X X X X
T 5 T 4 T 3 T 2 T 1 T
no.of ways 5! =120
Incorrect
5 places between the trees
0 0 0 0 0
X X X X X X
T 5 T 4 T 3 T 2 T 1 T
no.of ways 5! =120
Question 15 of 25
1 points
How many ways a team of 11 players can be made out of 15 players in one particular player is not to be selected in the team
Correct
$ifoneparticularplayerisnotselected\phantom{\rule{0ex}{0ex}}noofways=14{C}_{11}\phantom{\rule{0ex}{0ex}}=14{C}_{3}\phantom{\rule{0ex}{0ex}}=\frac{14x13x12}{3x2x1}\phantom{\rule{0ex}{0ex}}=364$
Incorrect
$ifoneparticularplayerisnotselected\phantom{\rule{0ex}{0ex}}noofways=14{C}_{11}\phantom{\rule{0ex}{0ex}}=14{C}_{3}\phantom{\rule{0ex}{0ex}}=\frac{14x13x12}{3x2x1}\phantom{\rule{0ex}{0ex}}=364$
Question 16 of 25
1 points
Find the number of arrangements of 5 things taken out of 12 things ,in which one particular thing must always be included
Question 17 of 25
1 points
Exactly 3 girls are to be selected from 5 girls and 3 boys.the probability of selecting 3 girls will be
Correct
$given,5girls+3boys\phantom{\rule{0ex}{0ex}}requiredexactly3girls\phantom{\rule{0ex}{0ex}}probabilityof3girls=\frac{5{C}_{3}}{8{C}_{3}}=10/56=5/28$
Incorrect
$given,5girls+3boys\phantom{\rule{0ex}{0ex}}requiredexactly3girls\phantom{\rule{0ex}{0ex}}probabilityof3girls=\frac{5{C}_{3}}{8{C}_{3}}=10/56=5/28$
Question 18 of 25
1 points
There are 12 questions to be answered to be Yes or No.how many ways can these be answered?
Correct
$no.ofways={n}^{r}={2}^{12}=4096$
Incorrect
$no.ofways={n}^{r}={2}^{12}=4096$
Question 19 of 25
1 points
how many permutations can be formed from the letters of the word DRAUGHT. if both vowels may not be separated ?
Correct
words DRAUGHT
if both vowels may not be separated this mean that both vowels come together
D R G H T A U
6 5 4 3 2 1
no.of ways if two vowels come together=6!x2!=720×2=1440
Incorrect
words DRAUGHT
if both vowels may not be separated this mean that both vowels come together
D R G H T A U
6 5 4 3 2 1
no.of ways if two vowels come together=6!x2!=720×2=1440
Question 20 of 25
1 points
$if13{C}_{6}+2.13{C}_{5}+13{C}_{4}=15{C}_{x}then,x=$
Correct
$if13{C}_{6}+2.13{C}_{5}+13{C}_{4}=15{C}_{x}\phantom{\rule{0ex}{0ex}}(13{C}_{6}+13{C}_{5})+(13{C}_{5}+13{C}_{4})=15{C}_{x}\phantom{\rule{0ex}{0ex}}14{C}_{6}+14{C}_{5}=15{C}_{x}\phantom{\rule{0ex}{0ex}}X=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}15{C}_{6}=15{C}_{x}$
Incorrect
$if13{C}_{6}+2.13{C}_{5}+13{C}_{4}=15{C}_{x}\phantom{\rule{0ex}{0ex}}(13{C}_{6}+13{C}_{5})+(13{C}_{5}+13{C}_{4})=15{C}_{x}\phantom{\rule{0ex}{0ex}}14{C}_{6}+14{C}_{5}=15{C}_{x}\phantom{\rule{0ex}{0ex}}X=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}15{C}_{6}=15{C}_{x}$
Question 21 of 25
1 points
number of ways of shaking hands in a group of 10 people shaking hands to each other are
Correct
$forshakinghands\phantom{\rule{0ex}{0ex}}noofways=n{C}_{2}\phantom{\rule{0ex}{0ex}}=10{C}_{2}\phantom{\rule{0ex}{0ex}}=10x9/2x1\phantom{\rule{0ex}{0ex}}=45$
Incorrect
$forshakinghands\phantom{\rule{0ex}{0ex}}noofways=n{C}_{2}\phantom{\rule{0ex}{0ex}}=10{C}_{2}\phantom{\rule{0ex}{0ex}}=10x9/2x1\phantom{\rule{0ex}{0ex}}=45$
Question 22 of 25
1 points
$if15{C}_{3r}=15{C}_{r+3}thenfindthevalueofr$
Correct
$given,15{C}_{3r}=15{C}_{r+3}\phantom{\rule{0ex}{0ex}}then,3r+r+3=15\phantom{\rule{0ex}{0ex}}4r=12\phantom{\rule{0ex}{0ex}}r=3$
Incorrect
$given,15{C}_{3r}=15{C}_{r+3}\phantom{\rule{0ex}{0ex}}then,3r+r+3=15\phantom{\rule{0ex}{0ex}}4r=12\phantom{\rule{0ex}{0ex}}r=3$
Question 23 of 25
1 points
how many different words can be formed with the letters of the words LIBERTY
Correct
no of different word can be formed from the letter of LIBERTY=7!=5040
Incorrect
no of different word can be formed from the letter of LIBERTY=7!=5040
Question 24 of 25
1 points
in how many ways can a family consist of 3 children here different birthday in a leap yr.
Correct
no of ways can a family consist of 3 children here different birthday in a leap yr= 366x365x364
Incorrect
no of ways can a family consist of 3 children here different birthday in a leap yr= 366x365x364
Question 25 of 25
1 points
A student has 3 books on computer, 3 books on economics and 5 books on commerce.if these books are to be arranged subject wise , then these can be placed on an shelf in the number of ways
Correct
no of ways = 3! 3! 5! 3!
=6x6x120x6=25920
Incorrect
no of ways = 3! 3! 5! 3!
=6x6x120x6=25920