Question

In how may ways can $6$ persons be chosen, among $4$ officers and $8$ jawans, so as to include at least one officer?

• A. $896$
• B. $780$
• C. $668$
• D. $448$

Answer 1

Answer: (A)

$896$

Number of persons to be chosen $=6$, officers $=4$, Jawans $=8$

Total number of ways to select $6$ persons in such a way so as to include at least $1$ officer $=1$ officer ouf of $4$ and $5$ jawans out of $8$ jawans $=\left[{}^4{C}_1{\times }^8{C}_5\right]$ or $2$ officers out of $4$ and $4$ jawans out of $8$ $=\left[{}^4{C}_2{\times }^8{C}_4\right]$ or $3$ officers out of $4$ and $3$ jawans out of $8$ $=[{{}^{4}c}_3\times {{}^{8}c}_3]$ or $4$ officers out of $4$ and $2$ jawans out of $8$ $=\left[{}^4{C}_4{\times }^8{C}_2\right]$

Thus the total number of ways to choose the persons $=\left[{}^4{C}_1{\times }^8{C}_5\right]+\left[{}^4{C}_2{\times }^8{C}_4\right]+\left[{}^4{C}_3{\times }^8{C}_3\right]+\left[{}^4{C}_4{\times }^8{C}_2\right]$

$=\left[\frac{4!}{1!3!}\times \frac{8!}{5!3!}\right]+\left[\frac{41}{2!2!}\times \frac{8!}{4!4!}\right]+\left[\frac{41}{3!1!}\times \frac{8!}{3!5!}\right]+\left[\frac{41}{4!0!}\times \frac{8!}{2!6!}\right]=896$