Question

Eighteen guests have to be seated half on each side of a long table. Four particular guests desire to sit on one particular side and three on the other side.Determine the number of ways in which the sitting arrangements can be made.

Or

A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if atleast one black ball is to be included in the draw?

Answer 1

As four particular gusts have seated on one side, so remaining five on that side can be selected out of eleve, in ${}^{11}{C}_5$ways.

Since seven particular guests already desire to sit either of the sides. So we have remaining guests to sit either side of the table.

Thus nine guests can be amanged in 9! ways.

Similarly, on another side, out of nine guests have seated so remaining six on that side can be selected out of six in ${}^6{C}_6$ ways .

Thus nine guests can be arranged in 9! ways .

$\therefore$ Total number of ways in which 18 persons can be seated ${}^{11}{C}_5{\times }^6{C}_6\times 9!$

$=\frac{11!}{5!6!}\times 9!\times 1\times 9!$

$=\frac{11!}{5!6!}\times 9!\times 9!$

Or

We have, two white balls, three black balls and four red balls .

Case:I When one black and two other balls are drawn, then

Number of ways =${}^3{C}_1{\times }^6{C}_2=3\times 15=45$

Case :II When two black and one other balls are drawn then ,

Number of ways =${}^3{C}_2{\times }^6{C}_1=3\times 6=18$

Case :III When three black balls are drawn, then,

Number of ways =${}^3{C}_2{\times }^6{C}_0=1\times 1=1$

Total number of ways =$45+18+1=64$