Question

There are 5 white, 4 yellow, 3 green, 2 blue and 1 red balls, All balls are identical except for the color. These are to be arranged in a line in 5 places. Final the number of distinct arrangements.
A.2322
B.5230
C.1244
D.2111

Answer 1

As we can see there are 5 places and 5 different types (colors) of balls, try finding all the different cases and how we can put them in these places. Here we have to take following cases:-
(i)All balls different color
(ii)2 same color balls and other 3 distinct.
(iii)2 pairs of different colors and the remaining 1 ball of different color
(iv)2 balls of the same color and a trial of different color.
(v)A trial of 1 color and 2 single, single balls of different color.
(vi)A quad of single color and the remaining single from any other color.
(vii)All five balls of same color given

Complete step-by-step answer:
Let's start with a solution by seeing all the different ways of choosing the balls to be placed in these 5 places.
(i)All the places have different color balls and arrange these in 5 places
= 5!
= 120

(ii)2 same color balls and other 3 distinct.
Now as we have 1 red ball
we cannot choose 2 from this, thus we can choose from 4 colors, given by
$=$ $^{\text{4}}{{\text{c}}_{\text{1}}}{{\text{ }\\!\\!\times\\!\\!\text{ }}^{\text{4}}}{{\text{c}}_{\text{3}}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{5!}}{\text{2!}}$
=960
(divide by 2!: as 2 balls are same)

(iii)2 pairs of different colors and the remaining 1 ball of different color given by,
$=$ $^{\text{4}}{{\text{c}}_{\text{2}}}{{\text{ }\\!\\!\times\\!\\!\text{ }}^{3}}{{\text{c}}_{1}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{5!}}{\text{2!}\times \text{2!}}$
=540
(iv)2 balls of same color and a trial of different color
As we have choice to choose the 3 balls from green, yellow and white after that we can choose 2 same color ball from other 3 colors (since 1 ball is of red color) given by,
$=$ ${{\left( ^{3}{{\text{c}}_{1}} \right)}^{\text{2}}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{5!}}{\text{2! }\\!\\!\times\\!\\!\text{ 3!}}$
=90
(v)A trial of 1 color and 2 single, single balls of different color, given by
$=$ $^{3}{{\text{c}}_{\text{1}}}{{\text{ }\\!\\!\times\\!\\!\text{ }}^{\text{4}}}{{\text{c}}_{2}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{5!}}{\text{3!}}$
=360
(vi)A quad of single color and the remaining single from any other color given by,
$=$ $^{2}{{\text{c}}_{\text{1}}}{{\text{ }\\!\\!\times\\!\\!\text{ }}^{\text{4}}}{{\text{c}}_{1}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{5!}}{\text{4!}}$
=40
(vii)All five balls of same color given by,
1
Now to final total distinct arrangements, first add all the above cases
(1)+ (2) + (3) + (4) + (5) + (6) + (7) = 120+960+540+90+360+40+1=2111

Answer is option (D).

Note: In solving this type of question we should form different cases based on the given conditions. We should remember permutations and combinations formulas.