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Class 10 Class 12
The letters of the word ‘RANDOM’ are arranged as in a dictionary. What is the rank of word ‘RANDOM’?

Numbers of letters in 'RANDOM' = (R → 1, A → 1, N → 1, D → 1, O → 1, M → 1) = 6
Number of words that begin with A = straight P presuperscript 1 subscript 1 cross times 5 factorial space equals space 1 cross times 120 space equals space 120
Number of words that begin with D = straight P presuperscript 1 subscript 1 cross times 5 factorial space equals space 1 space cross times space 120 space equals space 120
Number of words that begin with M = straight P presuperscript 1 subscript 1 cross times 5 factorial space equals space 1 cross times 120 equals 120
Number of words that begin with O = straight P presuperscript 1 subscript 1 cross times 5 factorial space equals space 1 cross times 120 equals 120
Number of words that begin with N = straight P presuperscript 1 subscript 1 cross times 5 factorial space equals space 1 cross times 120 equals 120
So far, we have formed 120 + 120 + 120 + 120 + 120 = 600 words.
Now words starting with R.
Number of words starting with RAD = 1 x 1 x 1 x 3! = 6.
Number of words starting with RAM = 1 x 1 x 1 x 3! = 6.
Total words                     = 600 + 12 = 612.
613th word is RANDMO
614th word is RANDOM
∴ Rank of the word random in a dictionary is 614. 

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A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

            Event 1: A coin is tossed and the outcomes recorded.
                    
            Number of outcomes open curly brackets straight H. space straight T close curly brackets space equals space 2

rightwards double arrow                                      m = 2

Event 2: The coin is tossed again and the outcomes recorded.

            Number of outcomes open curly brackets straight H comma space straight T close curly brackets space equals space 2

rightwards double arrow                                       n = 2

Event 3: The coin is tossed third time and the outcomes recorded.

           Number of outcomes open curly brackets straight H comma space straight T close curly brackets space equals space 2

rightwards double arrow                                           p = 2

∴  By fundamental principle of counting, the total number of outcomes recorded

                                                     = straight m cross times straight n cross times straight p space equals space 2 cross times 2 cross times 2 equals 8




   

548 Views

Given 5 flags of different colours, how many different signals can be generated if each signal requires use of 2 flags, one below the other?



Number of ways of finding a flag for place 1 = 5

rightwards double arrow                           m = 5

Number of remaining flags = 4

Number of ways of finding a flag for place 2 to complete the signal = 4

rightwards double arrow                     n = 4

∴ By fundamental principle of counting, the number of signals generated

                       = straight m cross times straight n equals 5 cross times 4 equals 20
991 Views

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed.




Number of digits available = 5

Number of places for the digits = 3.

Number of ways in which place (x) can be filled = 5

                           m = 5

Number of ways in which place (y) can be filled = 5    (∵  Repetition is allowed)

                            n = 5 

Number of ways in which place (z) can be filled = 5    (∵ Repetition is allowed)

                             p = 5

∴ By fundamental principle of counting, the number of 3-digit numbers formed.

                           = m x n x p = 5 x 5 x 5 = 125
458 Views

How many 3-digit odd numbers can be formed from the digits 1,2,3,4,5,6 if:
(a) the digits can be repeated (b) the digits cannot be repeated?



(a) Number of digits available = 6

Number of places [(x), (y) and (z)] for them = 3

Repetition is allowed and the 3-digit numbers formed are odd

Number of ways in which box (x) can be filled = 3 (by 1, 3 or 5 as the numbers formed are to be odd)

rightwards double arrow               m = 3
Number of ways of filling box (y) = 6                           (∴ Repetition is allowed)

rightwards double arrow               n = 6

Number of ways of filling box (z) = 6                           (∵ Repetition is allowed) 

rightwards double arrow              p = 6

∴  Total number of 3-digit odd numbers formed

                             = m x n x p = 3 x 6 x 6 = 108

(b) Number of ways of filling box (x) = 3                     (only odd numbers are to be in this box )  

rightwards double arrow                                   m = 3

Number of ways of filling box (y) = 5                                (∵ Repetition is not allowed)

rightwards double arrow                              n = 5

Number of ways of filling box (z) = 4                                 (∵ Repetition is not allowed)

rightwards double arrow                             p = 4

∴     Total number of 3-digit odd numbers formed

                                  = m x n x p = 3 x 5 x 4 = 60.   
231 Views

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed.


Number of ways in which place (x) can be filled = 5

                                           m = 5

Number of ways in which place (y) can be filled = 4      (∵ Repetition is not allowed)

                                           n = 4

Number of ways in which place (z) can be filled = 3      (∵ Repetition is not allowed)

                                           p = 3

∴ By fundamental principle of counting, the total number of 3 digit numbers formed

                                         = m x n x p = 5 x 4 x 3 = 60.

526 Views