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High School Mathematics - 2
13.2 Combinations and Permutations

Combinations

If n objects are given and we have to choose r (r < n) out of them and the order of choosing these r objects is not important, then such a choice is called a combination of n objects taken r at a time.

nCr = The number of combinations of 'n' things taken 'r' at a time.

Permutations

if n objects are given and we have to arrange r (r < n) out of them and the order of arranging these r objects is important then such a choice is called permutation of n objects taken r at a time

nPr = = The number of Permutations of 'n' things taken 'r' at a time.

A combination consisting of r objects can give rise to a number of permutation
eg:Q>Find the combination of 4 letters g ,a, t, e, taken 3 at a time
Solution
  1. g a t
  2. a t e
  3. t e g
  4. e g a

Consider 1.
This combination g a t can give six permutation
  • g a t
  • g t a
  • a g t
  • a t g
  • t g a
  • t a g
    So combinations 1-4 will give rise to 6x4 = 24 permutations

    Formulae

    • nCr = =n!/r!(n-r)!
    • nPr = =n!/(n-r)!

    Solved examples

    1. 4 persons enter a railway compartment in which there are 6 seats. In how many ways can they take their places

      Solution

      4 persons can take 6 seats in 6P4 ways
      ans = 360 ways

    2. How many 4 digit numbers can be formed out of the digits 2, 3, 4, 5, 6, 7 if no digit is repeated in any number? How many such numbers will be greater then 4000?

      Solution

      a) 4 digits can be arranged from 6 digits in 6P4 ways
      =360 ways
      b) Numbers greater than 4000.
      Thousands place can be filled in 4 ways (4,5,6,7)
      Having filled thousands place now 5 digits remain. So hundreds palce can be filled in 5P1 = 5 ways
      Having filled hundreds place now 4 digits remain. So tens place can be filled in 4P1 = 4ways
      Having filled tens place now 3 digits remain. So units place can be filled in 3P1 = 3 ways
      So the total number of ways in which we can get a number greater than 4000 is 4x5x4x3 = 240 ways

    3. From 12 books, in how many ways can a selection of 5 be made (a) When one specified book is always included (b)when one specified book is always excluded.

      Solution
      a) One book is always included. So 4 books can be chosen from the remaining 11 in 11C4 ways = 330 ways
      b) One book is always excluded. So 5 books can be chosen from the remaining 11 in 11P5 ways = 462 ways

    4. In a party of 15, each person shakes hands with every other person. What is the total number of handshake?

      Solution

      Consider the first person. He shakes hands with 14 other persons.
      1- 2,3,4,5,6,7,8,9,10,11,12,13,14,15 = 14 handshakes
      2- 1,3,4,5,6,7,8,9,10,11,12,13,14,15 . But 2 with 1 is already included above.
      So the total number of handshakes = 13
      3- 1,2,4,5,6,7,8,9,10,11,12,13,14,15. But 3 with 2 and 1 is already included
      So the total number of handshakes = 12
      so the actual number of handshakes = 14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105

    5. There are 8 different locks with exactly one key for each lock. All the keys have been mixed up. What is the maximum no of trials required to find out which key belongs to which lock.

      Solution

      Consider the first key. At the most all 7 trials will fail .Which implies that the key belongs to the eighth lock
      Consider the second key. At the most all 6 trials will fail .Which implies that the key belongs to the seventh lock
      So the maximum number of trials will be 7+6+5+4+3+2+1 = 28 trials.


Q 1: A combination can give rise to many permutations
yes
no

Q 2: The combination of 5 objects taken 3 at a time gives rise to how many permutations?
65
30
120
60

Q 3: In combination
the order of chosing the objects is important
the order of chosing the objects is not important

Q 4: Tossing of 2 coins give rise to how many combinations?
7
4
8
6

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Question 6: This question is available to subscribers only!


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