Question

(b). A club consists of 6 freshmen, 5 sophomores, and 4 juniors. How many committees of four people not from one class

Answer 1

To find the number of committees of four people not from one class, you can use combinations. There are 1365 possible committees of four people not from one class in the club.

Step-by-step explanation:

To find the number of committees of four people not from one class, you can use combinations. First, determine the number of ways to choose four people from the total number of students in the club, which is 6 freshmen + 5 sophomores + 4 juniors = 15 students.

This can be calculated as 15 choose 4, which is written as C(15, 4) or (15, 4). Use the formula:

(n, r) = n! / (r! * (n-r)!), where n is the total number of students and r is the number of students to choose. Plugging in the values, you get:

(15, 4) = 15! / (4! * (15-4)!)

= 15 * 14 * 13 * 12 / (4 * 3 * 2 * 1)

= 1365

So, there are 1365 possible committees of four people not from one class in the club.

Answer 2

Answer: 1344 committees can be formed of four people not from one class

Step-by-step explanation:

We are given:

Members in freshmen = 6

Members in sophomores = 6

Members in juniors = 6

To form a committee of four people not from one class, numerous combinations can occur. They are:

• When in a committee, at least one of them are from each class

When 2 from freshmen, 1 from sophomores, and 1 from junior. The possibility becomes
`^6C_2* ^5C_1* ^4C_1`

When 1 from freshmen, 2 from sophomores, and 1 from junior. The possibility becomes
`^6C_1* ^5C_2* ^4C_1`

When 1 from freshmen, 1 from sophomores, and 2 from junior. The possibility becomes
`^6C_1* ^5C_1* ^4C_2`

Total combinations:
`(^6C_2* ^5C_1* ^4C_1)+(^6C_1* ^5C_2* ^4C_1)+(^6C_1* ^5C_1* ^4C_2)=(300+240+180)=720`

• If two members come from same class

When 2 from freshmen, 2 from sophomores, and 0 from junior. The possibility becomes
`^6C_2* ^5C_2* ^4C_0`

When 0 from freshmen, 2 from sophomores, and 2 from junior. The possibility becomes
`^6C_0* ^5C_2* ^4C_2`

When 2 from freshmen, 0 from sophomores, and 2 from junior. The possibility becomes
`^6C_2* ^5C_0* ^4C_2`

Total combinations:
`(^6C_2* ^5C_2* ^4C_0)+(^6C_0* ^5C_2* ^4C_2)+(^6C_2* ^5C_0* ^4C_2)=(150+60+90)=300`

• If three members come from same class

When 3 from freshmen, 1 from sophomores, and 0 from junior. The possibility becomes
`^6C_3* ^5C_1* ^4C_0`

When 3 from freshmen, 0 from sophomores, and 1 from junior. The possibility becomes
`^6C_3* ^5C_0* ^4C_1`

When 0 from freshmen, 3 from sophomores, and 1 from junior. The possibility becomes
`^6C_0* ^5C_3* ^4C_1`

When 1 from freshmen, 3 from sophomores, and 0 from junior. The possibility becomes
`^6C_1* ^5C_3* ^4C_0`

When 0 from freshmen, 1 from sophomores, and 3 from junior. The possibility becomes
`^6C_0* ^5C_1* ^4C_3`

When 1 from freshmen, 0 from sophomores, and 3 from junior. The possibility becomes
`^6C_1* ^5C_0* ^4C_3`

Total combinations:
`[(^6C_3* ^5C_1* ^4C_0)+(^6C_3* ^5C_0* ^4C_1)+(^6C_0* ^5C_3* ^4C_1)+(^6C_1* ^5C_3* ^4C_0)+(^6C_0* ^5C_1* ^4C_3)+(^6C_1* ^5C_0* ^4C_3)]=(100+80+40+60+20+24)=324`

Total number of committees that can be formed = [720 + 300 + 324] = 1344

Hence, 1344 committees can be formed of four people not from one class