Difficult Math Problem #68

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800guy: Master | Next Rank: 500 Posts; Posts: 354; Joined: Tue Jun 27, 2006 9:20 pmFollowed by: 5 members; Thanked: 11 times

by 800guy » Fri Dec 01, 2006 5:27 pm

A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

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Source: Beat The GMAT — Problem Solving | Fri Dec 01, 2006 5:27 pm

kulksnikhil: Senior | Next Rank: 100 Posts; Posts: 58; Joined: Tue May 23, 2006 3:35 pm; Location: Seoul, Korea; Thanked: 2 times

Re: Difficult Math Problem #68

by kulksnikhil » Sat Dec 02, 2006 4:10 am

800guy wrote:A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

Person 1 can make team with 7 people
Person 2 can make team with 6 people (this excludes teaming with 1)
Person 3 can make team with 5 people (this excludes teaming with 1 and 2)
.
.
.
Peron 7 can make team with 1 people (excluding 1...6)

Total number of ways 2 member teams could be formed = 7!

Since we need groups of 4 teams...= 7!/4

But this is not in options... How to proceed ahead?

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800guy: Master | Next Rank: 500 Posts; Posts: 354; Joined: Tue Jun 27, 2006 9:20 pmFollowed by: 5 members; Thanked: 11 times

OA

by 800guy » Mon Dec 04, 2006 2:42 pm

OA:

1st team could be any of 2 guys... there would be 4 teams (a team of A&B is same as a team of B&A)... possible ways 8C2 / 4.
2nd team could be any of remaining 6 guys. There would be 3 teams (a team of A&B is same as a team of B&A)... possible ways 6C2 / 3
3rd team could be any of remaining 2 guys... there would be 2 teams (a team of A&B is same as a team of B&A). Possible ways 4C2 / 2
4th team could be any of remaining 2 guys... there would be 1 such teams... possible ways 2C2 / 1

total number of ways...

8C2*6C2*4C2*2C2
-------------------
4 * 3 * 2 * 1

=
8*7*6*5*4*3*2*1
--------------------
4*3*2*1*2*2*2*2

= 105 (ANSWER)...

Another method: say you have 8 people ABCDEFGH

now u can pair A with 7 others in 7 ways.
Remaining now 6 players.
Pick one and u can pair him with the remaining 5 in 5 ways.

Now you have 4 players.
Pick one and u can pair him with the remaining in 3 ways.

Now you have 2 players left. You can pair them in 1 way

so total ways is 7*5*3*1 = 105 ways i.e. E

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projase: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Sun Dec 03, 2006 2:45 pm

Re: OA

by projase » Wed Dec 06, 2006 2:34 pm

A question, why do you divide each combination by the number of groups?.

Regards,

PR

800guy wrote:OA:

1st team could be any of 2 guys... there would be 4 teams (a team of A&B is same as a team of B&A)... possible ways 8C2 / 4.
2nd team could be any of remaining 6 guys. There would be 3 teams (a team of A&B is same as a team of B&A)... possible ways 6C2 / 3
3rd team could be any of remaining 2 guys... there would be 2 teams (a team of A&B is same as a team of B&A). Possible ways 4C2 / 2
4th team could be any of remaining 2 guys... there would be 1 such teams... possible ways 2C2 / 1

total number of ways...

8C2*6C2*4C2*2C2
-------------------
4 * 3 * 2 * 1

=
8*7*6*5*4*3*2*1
--------------------
4*3*2*1*2*2*2*2

= 105 (ANSWER)...

Another method: say you have 8 people ABCDEFGH

now u can pair A with 7 others in 7 ways.
Remaining now 6 players.
Pick one and u can pair him with the remaining 5 in 5 ways.

Now you have 4 players.
Pick one and u can pair him with the remaining in 3 ways.

Now you have 2 players left. You can pair them in 1 way

so total ways is 7*5*3*1 = 105 ways i.e. E

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800guy: Master | Next Rank: 500 Posts; Posts: 354; Joined: Tue Jun 27, 2006 9:20 pmFollowed by: 5 members; Thanked: 11 times

by 800guy » Wed Dec 06, 2006 2:38 pm

for this type of question, you have to divide by the factorial of number of teams. i can't really explain, except that's how i've seen the formula

by the way, this isn't my answer, it's the OA i have

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projase: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Sun Dec 03, 2006 2:45 pm

by projase » Thu Dec 07, 2006 11:06 am

I think the asnwer is 2520, because for the first group, you select 2 people from a group of 8, ie 8 alternatives for the first person and 7 for the second. Because the order doesn't matter you divide by 2!, then 8C2 is already divided by the factorial number.

Do you know any reason to divide it again by 4?

Regards,

PR

800guy wrote:for this type of question, you have to divide by the factorial of number of teams. i can't really explain, except that's how i've seen the formula

by the way, this isn't my answer, it's the OA i have

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thankont: Senior | Next Rank: 100 Posts; Posts: 41; Joined: Fri Dec 15, 2006 9:23 amFollowed by: 1 member; Location: Greece

by thankont » Tue Dec 19, 2006 1:51 am

Another solution could be this (8C4*4C2)/4 (choose 4 people out of 8 and 2 out of 4 and then divide by number of teams) since we need 4 perons to play doubles and we choose them out of 8

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