A committee of 7 members is to be formed to put up the Christmas decor

Question

A committee of 7 members is to be formed to put up the Christmas decorations at Hogwarts. The following volunteered: 9 teachers, 7 Hufflepuff students, 4 ghosts, 6 Gryffindors, and 3 Prefects. How many possible committees can be formed by Professor Dumbledore if a committee must have only 1 Prefect, only 1 ghost, and at most 2 teachers?

A. 123,552
B. 200,772
C. 216,216
D. 316,188
E. 1,560,780

A. 123,552
B. 200,772
C. 216,216
D. 316,188
E. 1,560,780

sahuayush · Answer

Since the question asks the number of possible committees where order does not matter, it is a Combination problem.

As this questions pits a constraint of atmost 2 teachers, three cases can be identified-

1. 1P, 1G, 1T, 4 Students
2. 1P, 1G, 2T, 3 Students
3. 1P, 1G, 0T, 5 Students

Case 1-

3C1 X 4C1 X 9C1 X 13C4 ways

On calculating, this would be 77,220 ways

Similarly, case 2 would amount to-

3C1 X 4C1 X 9C2 X 13C3 ways = 123,552 ways

Case 3-

3C1 X 4C1 X 13C5= 15444 ways

Add all cases-

Total ways- 216,216 ways
Option C

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CrackverbalGMAT · Answer

Let Hufflepuff and Gryiffindor be put under a common category, Students = 6 + 7 = 13

When we say At most it means Maximum to Minimum (Including 0)

3 cases arise

1. 1P, 1G, 2T, 3 Students = 3C1 X 4C1 X 9C2 X 13C3 = 123,552 ways

2. 1P, 1G, 1T, 4 Students = 3C1 X 4C1 X 9C1 X 13C4 = 77,220 ways

3. 1P. 1G, 0T, 5 Students = 3C1 X 4C1 X 9C0 X 13C5 = 15,544 ways

Total = 123552 + 77220 + 15544 = 216,216

Option C

Arun Kumar

chetan2u · Answer

9 teachers, 7 Hufflepuff students, 4 ghosts, 6 Gryffindors, and 3 Prefects.

Ways to choose 
1) only 1 Prefect,---- 3 ways
2) only 1 ghost, ---- 4 ways
Now 5 remaining members
3) at most 2 teachers
a) no teacher =  13C5=\frac{13*12*11*10*9}{5*4*3*2}=13*11*9 
b) 1 teacher =  13C4*9C1=\frac{13*12*11*10}{4!*9}=13*11*5*9 
c) 2 teachers =  13C3*9C2=\frac{13*12*11}{3!}*\frac{9*8}{2}=13*11*9*8 
=>  13*11*9+13*11*9*5+13*11*9*8=13*11*9(1+5+8)=13*11*9*14

Total ways =  13*11*9*14*3*4 = 216,216

Note :  You could work on the units digit , and get the calculation 13*11*9*14*3*4 sorted out in 10 seconds.=> 3*9*4*3*4 = 9*9*4*4=1*6=6..Only C possible.
May not be everytime, but you can be lucky on many occasions.

ScottTargetTestPrep · Answer

Solution:

The committee can consist of: 1) 1 prefect, 1 ghost, 5 students; or 2) 1 prefect, 1 ghost, 1 teacher, 4 students; or 3) 1 prefect, 1 ghost, 2 teachers, 3 students.

Let’s calculate the number of committees for each of the cases.

1) 1 prefect, 1 ghost, 5 students

Notice that there are 7 + 6 = 13 students in total. 5 students can be chosen from among 13 students in 13C5 = 13!/(5!*8!) = (13*12*11*10*9)/(5*4*3*2) = 13 * 11 * 9 = 1,287 ways. Since there are three choices for the prefect and four choices for the ghost, there are 3 * 4 * 1,287 = 15,444 ways to form the committee with one prefect, one ghost and five students.

2) 1 prefect, 1 ghost, 1 teacher, 4 students

4 students can be chosen from among 13 students in 13C4 = 13!/(4!*9!) = (13*12*11*10)/(4*3*2) = 13 * 11 * 5 = 715 ways. Together with the 3 choices for the prefect, 4 choices for the ghost and 9 choices for the teacher, there are 3 * 4 * 9 * 715 = 77,220 ways to form the committee with one prefect, one ghost, one teacher and four students.

3) 1 prefect, 1 ghost, 2 teachers, 3 students

3 students can be chosen from among 13 students in 13C3 = 13!/(3!*10!) = (13*12*11)/(3*2) = 13 * 2 * 11 = 286 ways. 2 teachers can be chosen from among 9 teachers in 9C2 = 9!/(2!*7!) = (9*8)/2 = 36 ways. All together, there are 3 * 4 * 36 * 286 = 123,552 ways to form the committee with one prefect, one ghost, two teachers and three students.

Adding the number of ways for each case, we see that there are 15,444 + 77,220 + 123,552 = 216,216 ways to form the committee.

Answer: C

bumpbot · Answer

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A committee of 7 members is to be formed to put up the Christmas decor

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