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A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to :

 

  • Option 1)

    28

  • Option 2)

    27

  • Option 3)

    25

  • Option 4)

    24

 
atleast one boy & one girl :  ( 1B & 2G) + ( 2B & 1G)   As, n cannot be -ve so, n = 25 Option 1) 28 Option 2) 27 Option 3) 25 Option 4) 24

The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is: 

 

 

  • Option 1)

    2^{20}-1

  • Option 2)

    2^{20}+1

  • Option 3)

    2^{21}

  • Option 4)

    2^{20}

 
No. of ways Option 1) Option 2) Option 3) Option 4)

iIf three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is : 

  • Option 1)

    \frac{1}{5}

  • Option 2)

    \frac{3}{20}

  • Option 3)

    \frac{3}{10}

  • Option 4)

    \frac{1}{10}

 
Only two equilateral triangle are possible  and    Option 1) Option 2) Option 3) Option 4)

Suppose that 20 pillars of the same height have been erected along the boundary

of a circular stadium. If the top of each pillar has been connected by beams with the

top of all its non-adjacent pillars, then the total number of beams is : 

  • Option 1)

    170

  • Option 2)

    180

  • Option 3)

    210

  • Option 4)

    190

 
Any two non-adjacent pillars are joined by beams.  no. of beams = no. of diagonals                                                                                                             Option (1) is correct. Option 1) 170 Option 2) 180 Option 3) 210 Option 4) 190

Assume that each born child is equally likely to be a boy or a girl. If two 

families have two children each, then the conditional probability that

all children are girls given that at least two are girls is:

  • Option 1)

    \frac{1}{11}

  • Option 2)

    \frac{1}{10}

  • Option 3)

    \frac{1}{12}

  • Option 4)

    \frac{1}{17}

 
There are 4 children  total number of ways in whcih atleast 2 girls are there Required probabilty =  Option (1) is correct. Option 1) Option 2) Option 3) Option 4)

The number of 6 digit numbers that can be formed using the 

digits 0,1,2,5,7 and 9 which are divisible by 11 and no digit is repeated, is:

  • Option 1)

    72

  • Option 2)

    60

  • Option 3)

    48

  • Option 4)

    36

 
Sum of the given digit is 0+1+2+5+7+9 = 24 6 digit number let abcdef  abcdef is divisible by 11 if   is a multiple of 11  Only one possible is there a+c+e=b+d+f=12 Case 1: {a,c,e}={7,5,0}              {b,d,f}={9,2,1}  So, 2 x 2! x 3! = 24 Case 2: {a,c,e}={9 , 2, 1}              {b,d,f}={7, 5, 0} So, 3! x 3! = 36 Total = 24 + 36 = 60 So, correct option is (2). Option 1) 72 Option 2) 60 Option...

A committee of 11 members is to be formed from 8 males and 5 females. if m is the number of ways the commitee is formed with at least 6 males and n is the number of ways the commitee is formed with at least 3 females, then :

  • Option 1)

    m+n=68

  • Option 2)

    m=n=78

  • Option 3)

     n=m-8

  • Option 4)

    m=n=68

 
                                     Option 1) Option 2) Option 3)   Option 4)

The number of four - little number strictly greater than 4321 than can be formed using the digits 0,1,2,3,4,5(repetition of digits is allowed) is : 

 

 

  • Option 1)

    288

  • Option 2)

    360

  • Option 3)

    306

  • Option 4)

    310

 
(1)                   (2)                                                       (3)                                                               Option 1) Option 2) Option 3) Option 4)

All possible numbers are formed using the digits 1,1,2,2,2,2,3,4,4 taken all at a time. The number of such numbers in which the odd digits occupy even places is :

  • Option 1)

    180

  • Option 2)

    160

  • Option 3)

    175

  • Option 4)

    162

 
We have total 4 even place and 3 odd digit = Number of way placing odd digit at even places = Number of ways placing even digits So total number of ways = Option 1) Option 2) Option 3) Option 4)

There are m men and two women participating in a chess tournament. Each paricipant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is :

  • Option 1)

     

    9

  • Option 2)

     

    11

  • Option 3)

    12

  • Option 4)

    7

  Theorem of Combination - Each of the different groups or selection which can be made by taking r things from n things is called a combination. - wherein Where     Option 1)  Option 2)  Option 3)Option 4)

Consider three boxes, each containing 10 balls labelled 1,2,\cdots ,10. Suppose one ball is randomly drawn from each of the boxes. Denote by n_{i}, the label of the ball drawn from the i^{th} box, \left ( i=1,2,3 \right ). Then, the number of ways in which the balls can be chosen such that n_{1}< n_{2}< n_{3} is : 

  • Option 1)

     

    164

  • Option 2)

     

    120

  • Option 3)

     

    82

  • Option 4)

     

    240

  Theorem of Combination - Each of the different groups or selection which can be made by taking r things from n things is called a combination. - wherein Where    Option 1)  164Option 2)  120Option 3)  82Option 4)  240

The number of natural numbers less than 7,000 which can be formed by using the digits 0,1,3,7,9 (repitition of digits allowed) is equal to :

  • Option 1)

     

    250

  • Option 2)

     

    374

  • Option 3)

     

    372

  • Option 4)

     

    375

  Number of Permutations with repetition - The number of ways of filling r places where each place can be filled by any one of  n objects. It is  - wherein Where    Natural numbers less than 7,000 There are 3 ways to fill position 1 (i.e. 0,1,3)  and for the remaining position there are 5 ways each so, 0000 is not included.Option 1)  250Option 2)  374Option 3)  372Option 4)  375

Let S be the set of all triangles in the xy-plane,each having one vertex at the origin and the other two vertices  lie on coordinate axes with integral coordinates. If each triangles is S has area 50 sq. units, then the number of the elements in the set S is:

  • Option 1)

     

    9

  • Option 2)

     

    32

  • Option 3)

     

    18

  • Option 4)

     

    36

  Number of Divisions - The number of divisors of a natural number   is  - wherein Where a1, a2 ....... are distinct prime and non negative integers.     Let  and  be vectors of  Area of  le is  Number of triangles   Option 1)  9Option 2)  32Option 3)  18Option 4)  36

Consider a class of 5 girls and 6 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be members of the same team, is:

  • Option 1)

    500

  • Option 2)

    200

  • Option 3)

    300

  • Option 4)

    250

  Theorem of Combination - Each of the different groups or selection which can be made by taking r things from n things is called a combination. - wherein Where    from the concept of combination. Required np. of ways  =Total no. of ways - when A and B are always included.  Option 1)500Option 2)200Option 3)300Option 4)250
Engineering
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 If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is :

 

  • Option 1)

     46th

     

  • Option 2)

     59th

     

  • Option 3)

    52nd

  • Option 4)

     58th

 
As we have learned Rank of any Word - We arrange the words according to dictionary. Eq. for SUNIL        Rank is 95 - wherein                                                                               Tarting with A     4! / 2! = 12 words  Starting with L   ---->  4! = 24 words  Starting with m ----->  4! / 2! = 12 words  Starting with S ----> A ----> 3! / 2! = 3 words                     ...
Engineering
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Two numbers are selected at random (without replacement) from the first six positive integers. If X denotes the smaller of the two numbers, then the expectation of X, is :

  • Option 1)

    \frac{5}{3}

  • Option 2)

    \frac{14}{3}

  • Option 3)

    \frac{13}{3}

  • Option 4)

    \frac{7}{3}

 
Option 1) Option 2) Option 3) Option 4)
Engineering
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Let T_{n} be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If T_{n+1}-T_n=10 then the value of n is :

 

 

  • Option 1)

    8

  • Option 2)

    7

  • Option 3)

    5

  • Option 4)

    10

 
As we have learned Theorem of Combination - Each of the different groups or selection which can be made by taking r things from n things is called a combination. - wherein Where       no. of selection of 3 vertices out of n vertices                Option 1) 8 Option 2) 7 Option 3) 5 Option 4) 10
Engineering
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How many different words can be formed by jumbling the letters in the word  MISSISSIPPI  in which no two S are adjacent?

  • Option 1)

    7\cdot \, ^{6}C_{4}\cdot \; ^{8}C_{4}

  • Option 2)

    8\cdot \, ^{6}C_{4}\cdot \; ^{7}C_{4}

  • Option 3)

    6\cdot \, 7\cdot \; ^{8}C_{4}

  • Option 4)

    6\cdot \, 8\cdot \; ^{7}C_{4}

 
As we have learned Number of arrangement of like and alike objects - The number of arrangements that can be formed using n objects out of which r, q and p are identical objects then total number of arrangements Ex. ALLAHABAD -    Arrange M,I,I,I,P,P,I No. of ways =   Now arrange 4 'S' in the gaps =     Option 1) Option 2) Option 3) Option 4)
Engineering
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8 - digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do not occupy odd places, is :

  • Option 1)

    160

  • Option 2)

    120

  • Option 3)

    60

  • Option 4)

    48

 
As we have learned Number of arrangement of like and alike objects - The number of arrangements that can be formed using n objects out of which r, q and p are identical objects then total number of arrangements Ex. ALLAHABAD -    Odd nos - 1,1,3 EVEN nos . - 2,2,4,4 Case (1) ; odd places filled by 2,2,4,4  No. of ways =  Odd palces filled by 2,2,2,4 NO. of ways =  72+48 =...
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 The sum of the digits in the unit’s place of all the 4-digit numbers formed by using the numbers 3, 4, 5 and 6, without repetition,is :

  • Option 1)

    432

  • Option 2)

    108

  • Option 3)

    36

  • Option 4)

    18

 
Option 1) 432 Option 2) 108 Option 3) 36 Option 4) 18
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