Sunday 24 November 2013

Problems on Permutauions and Combinations



Q.1. In how many different ways can the letters of the word ‘JUDGE’ be arranged in such a way that the vowels always come together?
         a) 160    b) 120    c) 48     d) 124      e) 96
Ans: c
Explanation:
In the word ‘JUDGE’ the vowels are ‘UE’
When vowels are taken together it will be treated as one letter
Then the number ways of arrangement of ‘JDG(UE) = 4!
The two vowels can be arranged in 2 ways i.e. 2!
Therefore, the required number of ways = 4! × 2! = 24 × 2 = 48 ways
Q.2. How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
         a) 1024    b) 540     c) 5040    d) 2520     e) 400
Ans: c
Explanation:
The word ‘LOGARITHMS’ contain  10 letters
Number of 4- letter words, which can be formed from this 10 letters = 10P4 = 10 × 9 × 8 × 7
                                                                                                              = 5040 words
Q.3. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
           a) 5470    b) 5040     c) 86400    d) 11760     e) 2660
Ans: d
Explanation:
Required number of ways = 8C5 × 10C6
                                         = 8 × 7 × 10×  7 × 3 = 11760
Q. 4. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated?
             a) 5   b) 10     c) 15    d) 20     e) 25
Ans: d
Explanation:
Because the numbers should be divisible by 5, the unit’s digit of each numbers must be 5.
Such number of ways =1
The digit at the 10th place should be any of the remaining 5 digits = 5ways
Then the 100th digit should be any one from the remaining 4 digits = 4 ways
Therefore, the total number of numbers formed as such = 1 × 5 × 4 = 20
Q. 5. The value of 75P2 is
            a) 2550   b) 2775    c) 1555    d) 5550     e) None of these
Ans: d
Explanation:
nPr = n!/(n – r)!
75P2 = 75!/73! = 75 × 74 × 73!/73! = 75 × 74 = 5550
Q. 6. How many word can be formed by using all the letters of the word, 'ALLAHABAD' ?
             a) 2520   b) 3270    c) 3780    d) 1890     e) 7560
Ans: e
Explanation:
In the word 'ALLAHABAD’ there are 4A, 2L, 1H, 1B and 1D = 9
Required number of word = 9!/ 4! × 2! × 1! × 1! × 1! = 9  × 8 × 7 × 6 × 5 × 4!/4! × 2
                                           = 9  × 8 × 7 × 3 × 5 = 7560
Q. 7. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
              a) 71   b) 69   c) 64    d) 56     e) 48
Ans: c
Explanation:
Such a combination will be 1 Black + 2 Non-black or 2 Black + 1 Non-black  or 3 black.
          = (3C1 × 6C2) + (3C2 × 6C1) + 3C3
          = (3 × 6 × 5/2 × 1) + (3 × 2/2 × 1 × 6) + 1
          = 45 + 18 + 1 = 64
Q. 8. There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period?
             a) 3600   b) 2400   c) 3200    d) 2800     e) None of these
Ans: a
Explanation:
5 periods can be organized in 6P5 ways
Remaining period can be arranged in 5P1 ways
Therefore, the total number of ways = 6P5  × 5P1
                                                          = (6 × 5× 4 × 3 × 2) × 5 = 3600
Q. 9. How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?
          a) 1480   b) 720   c) 360    d) 1420     e) 1680
  
Ans: e
Explanation:
The first two places can be filled in only 1 way using 3 and 5.
Remaining 4 places can be filled with any of the 8 remaining digits in 8P4 ways = 8 × 7 × 6 × 5
                                                                              = 1680
Q. 10. An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?
             a) 64   b) 110   c) 100    d) 80     e) 70 
Ans: d
Explanation:
A chair can be arranged in 10 ways
A table can be arranged in 8 ways
Therefore, one chair and one table can be arranged in 10 × 8 = 80 ways.

Problems on LCM and HCF




Q.1. Two numbers are in the ratio 2 : 3. If their L.C.M. is 48. What is sum of the numbers?
a)       28       b) 40     c) 64      d) 42      e) 32
Ans: b
Explanation:
Let the two numbers be 2x and 3x respectively.
LCM of 2x and 3x = 6x
Given, 6x = 48
So, x = 8
The required sum = 2x + 3x = 16 + 24 = 40. Ans.
     Q.2. What is the greatest number of four digits which is divisible by 15, 25, 40 and 75 ?
             a) 9800     b) 9600     c) 9400       d) 9200       e) 9000
              Ans: b
             Explanation:
             The greatest 4- digit number = 9999
             LCM of 15, 25, 40 and 75 = 600
            The remainder obtained when 9999 is divided by 600 = 399
            Therefore, the required greatest number = 9999 – 399 = 9600 Ans.
     Q.3. Three numbers are in the ratio of 2 : 3 : 4 and their L.C.M. is 240. Their H.C.F. is:
           a) 40       b) 30        c) 20    d) 10      e) None of these
           Ans: c
            Explanation :
Let the numbers be 2x, 3x and 4x

LCM of 2x, 3x and 4x = 12x

=> 12x = 240

=> x = 24012 = 20

H.C.F of 2x, 3x and 4x = x = 20
     Q.4.  Find the minimum number of square tiles required to pave the floor of a room of 2m 50cm   long and 1m 50cm broad ?
a)      50      b) 750      c) 45       d) 15     e) None of these
          Ans: d
          Explanation:
  HCF of 250 cm and 150 cm is 50 cm, which is the side of the tile

So, the required number of tiles  =  (250 × 150) / (50 × 50)  =  15
  
Q.5. What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder?
     a) 1108     b) 1683      c) 2007         d) 3363    e) None of these
    Ans: b
   Explanation:
   Just see which of the given choices satisfy the given conditions

Take 3363. This is not even divisible by 9. Hence this is not the answer

Take 1108. This is not even divisible by 9. Hence this is not the answer

Take 2007. This is divisible by 9.

2007 ÷ 5 = 401, remainder = 2 . Hence this is not the answer

Take 1683. This is divisible by 9.

1683 ÷ 5 = 336, remainder = 3

1683 ÷ 6 = 280, remainder = 3

1683 ÷ 7 = 240, remainder = 3

1683 ÷ 8 = 210, remainder = 3

Hence 1683 is the answer

    Q.6. Find the HCF of   22×32, 2×34×7

  a) 128       b) 126       c) 146       d) 434    e) 148
  Ans: b
  Explanation:
  HCF is Highest common factor, so we need to get the common highest factors among given values. So we got
                      2 × 3×3 × 7 = 126
  Q.7.  Find the HCF of 54, 288, 360
   a) 18    b) 36      c) 54     d) 108    e) 72
   Ans: a
   Explanation:
   Let’s solve this question by factorization method.
54 =2×33,  288 = 25×32,  360 = 23×32×5

  So HCF will be minimum term present in all three, i.e.
2×32=18

  Q.8.  Reduce  368/575  to the lowest terms.

      a) 30/25      b)  28/29          c) 16/25     d) 11/12    e) 13/15
   Ans: c
   Explanation:
     We can do it easily by in two steps
     Step1: We get the HCF of 368 and 575 which is 23
     Step2: Divide both by 23, we will get the answer 16/25
Q. 9. Three numbers are in the ratio 1 : 2 : 3 and their H.C.F is 12. The numbers are
           a) 12, 24, 36   b) 5, 10, 15    c) 4, 8, 12      d) 6, 8, 13    e) 10, 20, 30
Ans: a
Explanation:
Numbers are 1 × 12, 2 × 12 and 3× 12
                     i.e. 12, 24 and 36
Q.10. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
           a) 4   b) 5    c) 7     d) 9       e) 13
Ans:
Explanation:
Required number = HCF of ( 91 – 43), (183 – 91)   and (183 – 43)
                             = HCF of 48, 92, 140
                             = 4