Question

7. Eleven students go to lunch. There are two circular tables in the dining hall, one can seat 7 people, the other can hold 4. In how many ways can they be seated

Answer 1

Answer:

239,580 ways of seating

Step-by-step explanation:

11 students will be divided into 2 groups. One group of 7 people and one group of 4 people. So first we need to find the number of ways of dividing 11 students into these 2 groups.

First group is of 7 people. We have to select 7 people out of 11. The order of selection does not matter so this is a combination problem. Selecting 7 people from 11 can be expressed as 11C7.

Formula for combination is:

$^{n}C_{r}=\frac{n!}{r!(n-r)!}$

For the given case this would be:

$^{11}C_{7}=\frac{11!}{7! \times 4!}=330$

So, there are 330 ways of selecting a group of 7 from 11 students. When these 7 students are selected the remaining 4 will go to the other group. So, we can say there are 330 ways to divide the 11 students in groups of 7 and 4. Note that if you start with group of 4 students, the answer will still the same because 11C4 is also equal to 330.

Next we have to arrange 7 students on a round table. The number of possible arrangements would be = (7 - 1)! = 6! = 720

Similarly, to arrange 4 people on a round table, the number of possible arrangements would be = (4 - 1)! = 3! = 6

Since, for each selection of the 330 groups, there are 720 + 6 possible seating arrangements, so the total number of possible seating arrangements would be:

330 ( 720 + 6) = 239,580 ways

Thus, there are 239,580 ways of seating 11 students.