Question

Groups each containing 3 boys are to be formed out of 5 boys A,B,C,D and E such that no one group contains both C and D together. What is the maximum number of such different groups?

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is C $7$

${5_C}_3$

Total number of arrangements, when any $3$ boys are selected out of $5=$ ${5_C}_3$.

Now, when groups contain both $C$ and $D$, then their selection is fixed and the remaining $1$ boy can be selected out of the remaining $3$ boys. It can be done in ${3_C}_1$ ways.

So, number of groups, when none contains both $C$ and $D=$ total number of arrangements-number of arrangements when group contains both $C$ and $D=$ ${5_C}_3$ – ${3_C}_1$ $=10–3=7$