Question

Two groups of players consist of $6$ and $8$ players. In how many ways can a team of $11$ players be selected from these two groups if at least $4$ players are to be included from the first group?

• A. $334$
• B. $344$
• C. $120$
• D. $168$

Answer 1

Answer: (B)

$344$

Group $1$ has $6$ players

Group $2$ has $8$ players

The possible cases are:

1) Selecting $4$ from group $1$ and $7$ from group $2$

Number of ways of selecting $4$ from $6$ players $={}^6{C}_4$

Number of ways of selecting $7$ from $8$ players $={}^8{C}_7$

The total number of ways $={}^6{C}_4\times {}^8{C}_7=120$

2) Selecting $5$ from group $1$ and $6$ from group $2$

Number of ways of selecting $5$ from $6$ players $={}^6{C}_5$

Number of ways of selecting $6$ from $8$ players $={}^8{C}_6$

The total number of ways $={}^6{C}_5\times {}^8{C}_6=168$

3) Selecting $6$ from group $1$ and $5$ from group $2$

Number of ways of selecting 6 from 6 players $={}^6{C}_6$

Number of ways of selecting 5 from 8 players $={}^8{C}_5$

The total number of ways $={}^6{C}_6\times {}^8{C}_5=56$

Hence, the number of ways $=120+168+56=344$