Combination problems

[onlyexternaluse]

1.) There are 9 staffs which would be assigned to 3 projects. 3 Staffs to each project and no one is assigned to multiple projects, how many different combanations of project are possible?

A) 252
B) 1680
C) 2340
D) 362880
E) 592704

2.) There are 10 boys. 6 boys will be assigned to a football team, which consists of 1 goal, 2 defense, 2 mid, 1 forward. Only 2 boys can play goal and cannot play other position. How many group is possible?

A) 60
B) 210
C) 2580
D) 3360
E) 151200

3.) Ben needs to form a committee of 3 from a group of 8 engineers. If 2 engineers are too inexperienced to serve together on the committee, how many different committee can Ben form?
A) 20
B) 30
C) 50
D) 56
E) 336

4.) There are 10 potential gift. Jay can buy only 2 gifts. How many different pairs of gift can Jay buy?
A) 10
B) 20
C) 45
D) 90
E) 200

[fibbonnaci]

Please post one question per thread. This will bring clarity into the posts.

[harsh.champ]

Please post one question per thread. This will bring clarity into the posts.

______________
Surely,it looks a bit cluttered right now.

As for the solution,
Ans2)As we know that only 2 boys (suppose B1 and B2) can play goal,only 1 of them can remain in the team while the other sits out.
Case 1:Suppose B1 is in the team.Then,the rest 5 places can be selected in (9-1)8C5 ways. [9-1 since B2 sits out.]
Case 2:Suppose B2 is in the team.Then,the rest 5 places can be selected in (9-1)8C5 ways. [9-1 since B1 sits out.]
Hence,total no. of groups = 2x8C5 ways.

[Considering the topic COMBINATION PROBLEMS , I am not considering the permutations.]

But my answer is coming to be 56x2=112ways which isn't any of the option choices.
_______________
Can anybody point out my mistake or is the question wrong?

[harsh.champ]

Ans 3) Now,this question falls on the similar lines as the above Q2.
Lets consider those 2 engineers to be E1 and E2.

Case 1 :E1 is present in the committee
Then,rest of the 2 members can be selected in 6C2. [6 since E2 can not be present in the committee]

Case 2 :E2 is present in the committee
Then,rest of the 2 members can be selected in 6C2. [6 since E1 can not be present in the committee]

Case 3:None of E1 and E2 is present in the committee
The 3 members can be selected in 6C3 ways.

Hence,total no. of ways = 6C2 + 6C2 + 6C3 = 15 + 15 + 20 = 50.

Hence, (C) would be the answer choice.

[harsh.champ]

Ans 4) This seems to me as a very easy question.
Can anybody spot any trap in the question??

As Jay can buy only 2 gifts,total no. of different pairs of gift = 10C2 = 45.

Hoping no trap is there in the question.

[thephoenix]

1.) There are 9 staffs which would be assigned to 3 projects. 3 Staffs to each project and no one is assigned to multiple projects, how many different combanations of project are possible?

A) 252
B) 1680
C) 2340
D) 362880
E) 592704

IMO E

first we must find out how many groups of 3 can be made out of 9
its nothing but 9C3=84

now this 84 groups of three staff each can be placed in 3 projects in
84*84*84=592704

[fibbonnaci]

I appeal to users not to encourage multiple questions in one thread. i have requested onlyexternaluse to repost the questions in one question/ thread format. Lets take the responsibility of keeping the forum clean.

[kvamsy]

1.) There are 9 staffs which would be assigned to 3 projects. 3 Staffs to each project and no one is assigned to multiple projects, how many different combanations of project are possible?

A) 252
B) 1680
C) 2340
D) 362880
E) 592704

2.) There are 10 boys. 6 boys will be assigned to a football team, which consists of 1 goal, 2 defense, 2 mid, 1 forward. Only 2 boys can play goal and cannot play other position. How many group is possible?

A) 60
B) 210
C) 2580
D) 3360
E) 151200

3.) Ben needs to form a committee of 3 from a group of 8 engineers. If 2 engineers are too inexperienced to serve together on the committee, how many different committee can Ben form?
A) 20
B) 30
C) 50
D) 56
E) 336

4.) There are 10 potential gift. Jay can buy only 2 gifts. How many different pairs of gift can Jay buy?
A) 10
B) 20
C) 45
D) 90
E) 200

Answer for 1st question:
As there are 3 teams with 3 different members can be selected in 9c3*6c3*3c3 - - - > 1680
let me know the OA for the 1st question

[kvamsy]

Please post one question per thread. This will bring clarity into the posts.

______________
Surely,it looks a bit cluttered right now.

As for the solution,
Ans2)As we know that only 2 boys (suppose B1 and B2) can play goal,only 1 of them can remain in the team while the other sits out.
Case 1:Suppose B1 is in the team.Then,the rest 5 places can be selected in (9-1)8C5 ways. [9-1 since B2 sits out.]
Case 2:Suppose B2 is in the team.Then,the rest 5 places can be selected in (9-1)8C5 ways. [9-1 since B1 sits out.]
Hence,total no. of groups = 2x8C5 ways.

[Considering the topic COMBINATION PROBLEMS , I am not considering the permutations.]

But my answer is coming to be 56x2=112ways which isn't any of the option choices.
_______________
Can anybody point out my mistake or is the question wrong?

I too got the same thing. OA please