Break down the question to one choice at a time.

1st project 3 people. How many ways to choose 3 people out of 7?

7 possible options for the first choice
6 for the second
5 for the last.

Order of choosing doesn;t matter - we only want to know who we chose, not in which order we chose them. divide by number of choices factorial (3!) to get 7*6*5/3!

In other words: 7C3

2nd project 2 people. Hoe many ways to choose 2 people out of the remaining 4?
4*3/2!
Or in other words 4C2.

last project has only two people remaining, so number of choices is 2*1/2! = 1 (which makes sense, as there's only one way to choose 2 people out of 2, order doesn't matter).

Final answer is 7*6*5/3! * 4*3/2! * 1 = 7*5*2*3 = 210.

mariah wrote:A professor assigns three projects to seven students, therefore, the students will be divided two three groups, which has three, two, two students, respectively. How many ways are possible?

let students be a,b,c,d,e,f,g. out of these 7 students 3 students to form the group of 3 can be selected in 7C3 ways.!!

now from the remaining four students two groups of two students each can be formed in the 4!/2!*(2!)^2=3

hence total no. of ways would be 7C3 * 3 = 105

Hi Geva,

the number of ways of selection 3 out of 7 students is = 7C3
the number of ways of selection 1 out of 3 projects is = 3C1

Shouldn't it be 7c3*3c1 ??? question stem never said that students have fixed projects??

Similarly 4c2*2c1 ... last one 1

HSPA wrote:Hi Geva,

the number of ways of selection 3 out of 7 students is = 7C3
the number of ways of selection 1 out of 3 projects is = 3C1

Shouldn't it be 7c3*3c1 ??? question stem never said that students have fixed projects??

Not fixed project, on the contrary - yo're trying to count the 210 different ways of assigning the students to different projects.

Within the 7C3 =35 different ways of choosing the 3 people out of 7 to join the big project, you're counting such combinations as

ABC
ABD
ABE
...
BCD
BCE
BCF

etc.
There are 7 different people you can choose for the first slot: A, B, C, D, E, F, G.
Once you have chosen a person for the first slot, there are 6 people remaining. If we chose A for the 1st, we have B, C, D E, F, G.

And once we choose someone for THAT slot, we only have 5 people for the last slot.

Among all of these combinations 7*6*5 (divide by 3! because the order doesn't matter), you are counting all the different ways of assigning all combinations of all 7 members for the big project. The remaining 4C2 is the step where you count the different assignments of the remaining tema members to the next project, and the last project is already determined by these choices.

Geva@MasterGMAT wrote:

HSPA wrote:Hi Geva,

the number of ways of selection 3 out of 7 students is = 7C3
the number of ways of selection 1 out of 3 projects is = 3C1

Shouldn't it be 7c3*3c1 ??? question stem never said that students have fixed projects??

Not fixed project, on the contrary - yo're trying to count the 210 different ways of assigning the students to different projects.

Within the 7C3 =35 different ways of choosing the 3 people out of 7 to join the big project, you're counting such combinations as

ABC
ABD
ABE
...
BCD
BCE
BCF

etc.
There are 7 different people you can choose for the first slot: A, B, C, D, E, F, G.
Once you have chosen a person for the first slot, there are 6 people remaining. If we chose A for the 1st, we have B, C, D E, F, G.

And once we choose someone for THAT slot, we only have 5 people for the last slot.

Among all of these combinations 7*6*5 (divide by 3! because the order doesn't matter), you are counting all the different ways of assigning all combinations of all 7 members for the big project. The remaining 4C2 is the step where you count the different assignments of the remaining tema members to the next project, and the last project is already determined by these choices.

hi geva , lets assume that the last 4 people be a,b,c,d; now the possible pair of groups that can be formed is 4C2:
ab,cd
ac,bd
ad,bc
bc,ad
bd,ac
cd,ad

now if we see these groups, we will notice that last three group are same as the first three groups so i believe total no. of ways for selection would be half of 4C2 i.e. 3

manpsingh87 wrote:

Geva@MasterGMAT wrote:

HSPA wrote:Hi Geva,

the number of ways of selection 3 out of 7 students is = 7C3
the number of ways of selection 1 out of 3 projects is = 3C1

Shouldn't it be 7c3*3c1 ??? question stem never said that students have fixed projects??

Not fixed project, on the contrary - yo're trying to count the 210 different ways of assigning the students to different projects.

Within the 7C3 =35 different ways of choosing the 3 people out of 7 to join the big project, you're counting such combinations as

ABC
ABD
ABE
...
BCD
BCE
BCF

etc.
There are 7 different people you can choose for the first slot: A, B, C, D, E, F, G.
Once you have chosen a person for the first slot, there are 6 people remaining. If we chose A for the 1st, we have B, C, D E, F, G.

And once we choose someone for THAT slot, we only have 5 people for the last slot.

Among all of these combinations 7*6*5 (divide by 3! because the order doesn't matter), you are counting all the different ways of assigning all combinations of all 7 members for the big project. The remaining 4C2 is the step where you count the different assignments of the remaining tema members to the next project, and the last project is already determined by these choices.

hi geva , lets assume that the last 4 people be a,b,c,d; now the possible pair of groups that can be formed is 4C2:
ab,cd
ac,bd
ad,bc
bc,ad
bd,ac
cd,ad

now if we see these groups, we will notice that last three group are same as the first three groups so i believe total no. of ways for selection would be half of 4C2 i.e. 3

Nope, ab,cd is not the same as cd,ab, because the projects are different: In the first case a and b work on project nova while c and d work on project gemini, in the latter the roles are reversed. the order of choosing the teams within each project does not matter: If we choose ab or ba for the first project, the two are the same: in both cases we chose a and b for the project, so there;s no reason to count them as two different assignments (hence the division of each project by k!). But between different projects, the order matters because the projects are different groups.

shouldt it be like this?!
7c3*4c2*2c2 + 7c2*5c3*2c2 + 7c2*5c2*4c3=

hi my take is a little different. i agree that the total number of ways projects can be assigned to students is 105 (7c3 * 3)
however since we have further 3 projects which can further be arranged in 3! ways
hence total ways

105*3!= 630

mariah what is the OA?