Permutations And Combinations

Question
CBSEENMA11014162

In how many ways can 4 red, 3 yellow and 2 green balls be arranged in a row if the balls of the same colour is indistinguishable?

Solution

Total number of balls = space space space space 4 to the power of straight R plus 3 to the power of straight Y plus 2 to the power of straight G space equals space 9
 
rightwards double arrow                           n = 9
Number of red balls = 4
rightwards double arrow                    p = 4
       Number of yellow balls = 3
 rightwards double arrow                    q = 3
       Number of green balls = 2
rightwards double arrow                     r = 2
∴  The number of permutations of balls:
                       equals fraction numerator straight n factorial over denominator straight p factorial space straight q factorial space straight r factorial end fraction equals fraction numerator 9 factorial over denominator 4 factorial space 3 factorial space 2 factorial end fraction space equals space fraction numerator 9 cross times 8 cross times 7 cross times 6 cross times 5 cross times 4 cross times 3 cross times 2 cross times 1 over denominator 4 cross times 3 cross times 2 cross times 1 cross times 3 cross times 2 cross times 1 cross times 2 cross times 1 end fraction
                        = 9 x 4 x 7 x 5 = 1260.

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