Question

A box has 18 balls, 6 white balls numbered 1 to 6,6 black balls numbered 1 to 6 and 6 red balls numbered 1 to 6 . We have to divide all 18 balls into 6 groups (groups are not numbered) such that each group consists of one white, one black and one red ball and difference of numbers on any two balls of same group is at most one. Find no. of ways this division can be done.

• A. 97
• B. 98
• C. 99
• D. 100

Answer 1

Answer: (A)

Let $a_n$ be total ways of division when there are $\mathrm{n}$ balls of each colour.
Case I If we make one group $\mathrm{B}_1 \mathrm{~W}_1 \mathrm{R}_1$ remaining balls division can be done in $\mathrm{a}_{\mathrm{e}-1}$ ways.
Case II If we make one group $B_1 W_1 R_2$ remaining balls division can be done in $a_{-2}$ ways.
Case III If we make one group $B_1 W_2 R_1$ remaining balls division can be done in $\mathrm{a}_{\mathrm{u}-2}$ ways.
Case IV If we make one group $B_1 W_2 R_2$ remaining balls division can be done in $\mathrm{a}_{\mathrm{a}-2}$ ways.
$ \therefore a_n=a_{n-1}+3 a_{n-2} $
$ \text { Now } a_1=1, a_2=4 $
$ \therefore a_3=4+3=7 $
$ a_4=7+12=19 $
$ a_5=19+21=40 $
$ a_6=40+57=97$