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There are $5$ white,$4$ yellow,$3$ green,$2$blue and $1$ red ball.The balls are all identical except for colour. These are to be arranged in a line in $5$ places.Find the number of distinct arrangements.


There are following ways to arrange them in $4$ places
$(1)$ all different
$(2)$a pair and $3$ different
$(3)$two pair of different colours and $1$ different colour
$(4)$ a pair and a triad of different colours
$(5)$a triad of one colour and two singles of different colours
$(6)$a quad of one colour and a single of another colour
$(7)$ all five balls of same colour

Further i am stuck.The answer given in the book is $2111.$

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  • $\begingroup$ What do yo mean "arranged in a line in 5 places"? $\endgroup$ Commented Apr 6, 2016 at 10:40
  • $\begingroup$ It means $5$ balls in a line. $\endgroup$ Commented Apr 6, 2016 at 10:41
  • $\begingroup$ I think you are right - you just have to slog away and calculate each of (1)-(7). Most are fairly easy, but care is needed. $\endgroup$
    – almagest
    Commented Apr 6, 2016 at 11:09
  • $\begingroup$ Please add your calculations for the cases and indicate where you are stuck. $\endgroup$ Commented Apr 6, 2016 at 11:31

1 Answer 1

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"$(1)$ all different
$(2)$a pair and $3$ different
$(3)$two pair of different colours and $1$ different colour
$(4)$ a pair and a triad of different colours
$(5)$a triad of one colour and two singles of different colours
$(6)$a quad of one colour and a single of another colour
$(7)$ all five balls of same colour

Further i am stuck.The answer given in the book is $2111.$"

$1.$ $5! = 120$

$2.$ $C(4,1)C(4,3)5!/2! = 960 $

$3.$ $C(4,2)C(3,1)5!/(2!2!) = 540 $

$4.$ $C(3,1)C(3,1)5!/(2!3!) = 90 $

$5.$ $C(3,1)C(4,2)5!/3! = 360 $

$6.$ $C(2,1)C(4,1)5!/4! = 40 $

$7.$ $1$

It sums to $2111$.

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