Good Set of Problems in Modern Math (Solved)



  • Credits - Rajendra Rajput

    In how many ways can 8 men and 4 women be arranged in a row so that no two women are together?

    -M-M-M-M-M-M-M-M-
    9 gaps can be filled in 9c4 ways
    8 men can be arranged in 8! Ways
    4 women can be arranged in 4! Ways
    Total ways = 9c4 * 8! * 4!
    = 9! * 8 * 7 * 6
    = 121927680 ways

    Using the digit 2,2,2,3,3,4 . Find the sum of all six digit number that can be formed? (repitition not allowed)

    (N-1)! * (sum of digit) * (11111....[N-1] times)
    = 5! * 16 * 11111
    But since there is a repetition of three 2's ans two 3's
    = (5! * 16 * 11111)/(3! * 2!)
    = 120 * 16 * 11111/ 12
    = 10 * 16 * 11111
    = 1777760 Answer

    In a Family of 3 children, What is the probability of having atleast 1 boy?

    Probability of atleast 1 boy = 1 - probability of girl
    = 1 - (1/2)^3
    = 1 - 1/8
    = 7/8

    How may 4 digit numbers are there when a digit may be repeated any number of times in each numbers?

    1st place can be filled with any of the 9 numbers I.e (1-9)
    Rest 3 place place can be filled with any of the 10 numbers I.e (0-9)
    So answer is
    9 * 10 * 10 * 10
    I.e 9 * 10^3
    = 9000

    Or

    9 * 10^(n-1)
    = 9 * 10^(4-1)
    = 9 * 10^3
    = 9000

    A four-digit number is formed using the digits {0,6,7,8,9}. How many of these numbers are divisible by 3? (Each digit occurs atmost once in every no.)

    If sum of digit is divisible by 3 then the no. Is divisible by 3 .

    Only 3 such possibility here
    {6,7,8,0} ---> (3!*3) -->18 ways
    {7,8,9,0} ---> (3!*3) -->18 ways
    {6,7,8,9} ---> (4!) ---->24 ways

    Total ways = 18+18+24 = 60 ways

    In how many ways 20 beads can be arranged in a necklace such that 2 beads always stay together?

    20 beads
    2 always stay together
    So consider it a one entity
    Now total 19 beads
    Its should be in a necklace form
    So it will arrange in (n-1)! Ways
    (19-1)! = 18!
    Now (18!/2)*2! Because those 2 beads can arrange itself in 2! Ways
    So total 18! Ways

    How many permutations are there of the letters of the word MISSISSIPPI in which no four “I”s come together?

    MISSISSIPPI < --- 11 leters
    Repitition
    4 times S
    4 times I
    2 timee P

    Total ways to be arranged all letter is
    = permutation/repetition
    = 11!/(4!*4!*2!)
    =34650 ways

    Total ways when all I's come together
    Consider all the 4 I's as one entity
    (IIII) MSSPPSS
    Now total ways are
    = 8!/4!*2!
    = 840 ways

    Total ways when all I's are not together
    = 34650 - 840
    = 33810 ways

    The letters of word hostel are arranged so that vowels accept the end places. Then the number of arrangements?

    HOSTEL
    vowel -> O, E
    Consonant-> HSTL
    Fix the vowel at end place
    O - - - - E
    Consonant can be arranged in 4! Ways
    O & E can alter their position itself
    So total ways
    4! * 2
    = 24 * 2
    = 48 ways

    If any 3-digit number is picked at random, what is the probability that the number or its permutation (also a three digit no) is divisible by 5 and 4?

    Total
    9 * 10 * 10
    = 900 ways

    3 digit Number to be divisible by 5 is last digit must be either 5 or 0
    3 digit number to be divisible by 4 is last digit and tens must be 0,2,4,6,8
    Common is 0
    So in last place 0 will come
    _ _ 0
    Now tens place can be filled by 0,2,4,6,8 total 5 number
    Hundred place can be filled by any of the 9 number i.e 1,2,3.....9
    _ _ 0
    9c1 * 5c1 * 1
    45 ways

    Total probability
    45/900
    = 1/20
    0.05 ways

    There are 6 circles and 5 straight lines on a plane . Find the maximum number of points of intersection ?

    6 circle = m
    5 line = n
    nC2 + (mC2 * 2) + (n * m * 2)
    = 5C2 + (6C2 * 2) + (5 * 6 * 2)
    = 10 + 30 + 60
    = 100 max intersection

    The letter in the word ADOPTS are permuted in all possible ways and arranged in alphabetical order. How do you find out the word at position 42 in the permuted alphabetical order?

    In alphabetical order : A D O P S T
    A _ _ _ _ _ : the places filled in 5! ways = 120, But we need a rank less than 120. So the word starts with A.
    A D _ _ _ _ : empty places can be filled in 4!=24
    A O _ _ _ _ : the places filled with 4! ways = 24. If we add 24 + 24 this total crosses 42. So We should not consider all the words starting with AO.
    A O D _ _ _ : 3!= 6
    A O P _ _ _ : 3!=6
    Till this 24+12=36 words are obtained, we need the 42nd word.
    AOS _ _ _ : 3!= 6
    Exactly we are getting the sum 42. So last 3 letters in the descending order are TPD.
    So given word is AOSTPD

    Two cards are randomly chosen from a pack of 52 cards. what is the probability that one is jack and the other is black?

    There are 2 possibilities :-

    1. Red jack and black card
      2c1 * 26c1/52c2
      = 2 * 26/1326
      =52/1326

    2. black jack and black card
      2c1 * 25c1/52c2 [ 25 black card remains because out of 26 black card one of black jack is selected ]
      =2 * 25/1326
      =50/1326

    Total
    =52/1326 + 50/1326
    =102/1326
    =17/221

    What is the probability that 3 numbers chosen from 1 to 30 are consecutive?

    (1,2,3)(2,3,4).... (28,29,30)
    There are total 28 sets
    1 set is selected in 28C1 ways
    =28
    Total outcomes is 30C3
    = 30 * 29 * 28 / 3 * 2 * 1
    = 4060
    Probability = 28/4060
    = 1/145

    What is the probability of getting a sum of 4 in a toss of a pair of dice?

    (2,2) 1 ways
    (3,1) 2! Ways
    Total outcome 6^n
    6^2
    36

    Total probability
    1+2!/36
    1+2/36
    3/36
    1/12

    What is the probability of having a sum of 10 after rolling 3 dice?

    There are 6^3 possible outcomes to rolling a die 3 times. Out of these, how many yield a total of (exactly) 10 dots?
    First we find all sets {a,b,c} such that a+b+c=10
    1, 3, 6
    1, 4, 5
    2, 3, 5
    2, 2, 6
    2, 4, 4
    3, 3, 4
    The total number of sets that fit these criteria is 6. If a≠b≠c, then there exist 3! unique permutations of {a,b,c}. If a=b≠c, then there exist 3 unique permutations of {a,b,c}- There cannot be a set such that a=b=c
    There are 3 sets of the first kind and 3 of the second. It follow that the total number of triple die rolls that can fit the criteria is

    3 * 3! + 3 * 3
    18 + 9
    27
    So
    27/216
    1/8

    If a craftsman has 6 different seashells, how many different bracelets can be constructed if only four shells are to be used in any one bracelet?

    Number of ways of selecting 4 seashells=6C4

    Bracelet is in circular shape
    So circular and anti-circular arrangements are possible
    So arrangements will be (n-1)! Ways

    6C4 * (4-1)!
    = 30 * 3!
    = 180 bracelets

    How many passwords can be formed where 2 letters of the alphabet can be used, and one letter occurs twice and the other letter occurs three times? For example, the five characters in the password maybe be aadad or dadaa.

    1st letter can be selected in 26C1 ways = 26
    Now for 2nd letter we have to select from remaining 25 alphabet and this can be done in 25C1 ways = 25

    One letter occur twice and other letter occur thrice
    So a combination of 5 letter
    From the example given "aadad"
    It can be arranged in
    Total letter!/repetition!
    5!/(3!*2!) Ways

    So our answer is
    26 * 25 * 5!/(3!*2!)
    = 6500 ways

    So a total of 6500 password can be formed


Log in to reply
 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.