Question

A lady gives a dinner party for six guests. The number of ways in which they may be selected from ten friends, if two of the friends will not attend the party together, is?

• A. $112$
• B. $140$
• C. $164$
• D. $Noneofthese$

Answer 1

Answer: (B)

$140$

dinner party to six gents

no of ways in which they be selected $=?$

Total no. of ways to select $6$ guests from $10$ friends

${=}^{10}{C}_6$

$=\frac{10!}{6!4!}$

$=\frac{10\times 9\times 8\times 7}{4\times 3\times 2}$

$=210$

Total no. of ways of selecting the $2$ friends always

${=}^2{C}_2{\times }^8{C}_4=1\times \frac{8!}{4!4!}$

$=\frac{8\times 7\times 6\times 5}{4\times 3\times 2}=70$

Required no. of ways $=210-70=140$