A committee that includes 6 members is about to be divided into 2 subc

Question

A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member?

A. 10%
B. 20%
C. 25%
D. 40%
E. 50%

Scuven · Answer

I would say -> consider total possible cases as C(6,3) = 20

Consider favorable cases as C(5,2) =10 (considering 2 guys as one and total guys number as 5)

Thus p=10/20 =50%

BlackHawk2478 · Answer

Break up this seemingly hard question.
First of all, let's calculate the number of subcommittees of which Michael is a member. Reduce this problem into an easier form: assume that Michael is in one subcommittee, and find the number of ways to pick the two other members of the subcommittee, with no importance to order. Use the Combinations formula to pick 2 out of the 5 possible members, not including Michael:

Now, we want the number of subcommittees that include both Michael and David. Reduce this problem into an easier form: assume that both Michael and David are in one subcommittee and pick the third member out of the remaining 4 candidates. Don't formulate here: 1 guy out of 4—that's 4 options.

Alternative Method:
Visualize the problem thus: put 3 people in room A and the other 3 people in room B. Say Michael is in room A. Thus room A has 2 more open places out of 5 (the 2 places in Michael's room plus the 3 places in the other room). The chance that David will be in one of the 2 places in Michael's room is
=40%

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jack0997 · Answer

Bunuel

Could you help me find my mistake? I did this way and got the answer 50%.

Either Michael and David will be in the same team or not. So chances = 50%.

ntngocanh19 · Answer

I will try to help you,
1.Your total posible case is double because when you choose 3 people into a subcommittee, other three will form other subcommittee.
2. When you consider 2 guys as one, only four guy left to choose.
3. The Subcommittee includes 3 people so when you consider M&M as one, you need to choose 1 more person in 4 other people to join the subcommittee => It should be C(4,1)=4
The probability that M and M in a subcommittee will be 4/10 = 40%

Hope it helps

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ZulfiquarA · Answer

Michael is a part of subcommittee and we have to find what are the possibilities (probability) of David to be a part of it.
Since Michael is already in the committee, there are 5 people left and we have to find the probability of David being chosen.
Probability of David being chosen=1/5
There are two subcommittees so the probability doubles=1/5*2=2/5=.40 = .40*100=40%.

sujoykrdatta · Answer

We need to determine the number of cases when Michael (M) and David (D) are in the same group
If M and D are already in the group, we need to select one more member from the remaining 4. This can be done in 4C1 = 4 ways 
Note: the other 3 members will anyways form the other group
Thus, favorable cases = 4
Total cases: We need to select 3 from the 6 in 6C3 = 20 ways. However, this gives us one committee while the other is automatically selected when we have 3 members in the first. Thus, the number of ways is 20/2 = 10
 For example, if A, B, C, D, E, F are the members, 6C3 would include A-B-C and also C-D-E as committees. However, we don't need to count both since one implies the other.

Required percentage = 4/10 * 100 = 40%

Answer D

Regor60 · Answer

How many ways can the 6 people be distributed among 2 distinct teams ?:

6×5×4 ×3×2×1
_ _ _ _ _ _

3×2×1 ×3×2×1

= 20

The denominators adjust for the the fact that the order of three people in a team doesn't matter because they don't have unique roles within the team.

How many ways can the people be distributed if the two people are on the first team ?

M D 4 × 3 × 2 × 1
_ _ _ _ _ _
 3 × 2 × 1

= 4

But, there are two distinct teams, so the total number of ways equals

8

Percentage of teams in which Matthew and David are on the same team equals:

8/20 =40%

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mike12543 · Answer

lets just put these 6 members in a queue, 
0 0 0 0 0 0
 and we make the rule: these members have to split in the middle,
so that first three and last three are different subcommittees(making a rule does not change anything) 
0 0 0 / 0 0 0
so accroding to the question, Matthew and David have to be on the same side.
lets consider it in the opposite way, they dont meet each other in the same subcommittee,(it's the easy way to do)
the total combination is
6*5*4*3*2*1=720, (the queue)
we want Matthew and David on the different side:
now say Matthew is the left side,he has 3 possible positions; David on the right side has 3 possible positions. and they two can switch sides.
3*3*2=18
fill in with the other four guys.
4*3*2*1=24
so we have:
18*24=432, the possible combinations of them not being on the same side, 
the chance is 432/720=60%
so we have 1-60%=40%

Giyo · Answer

Please bunel explain this question clearly

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hughng92 · Answer

I agree with the 40% answer choice as that was what LBS Mock Test proposed as a correct answer choice. The main take-away is that the problem does not explicitly indicate whether the position of Michael and David matters within a subcommittee, hence it is quite simpler compared to other combinatorics problems.

Step 1:  Determine the denominator
How many ways to pick three people to form a subcommittee? C(6,3) = 20

Step 2:  Now you form two sub-committees, let's find out the how many ways you can form such sub-committees in which Michael and David belong together in one of them

Option 1: Sub-committee 1:  David – Michael –   Third Person 
How many ways can you choose the third person? C(4,1) = 4

Option 2: Sub-committee 2:  David – Michael –   Third Person 
How many ways can you choose the third person? C(4,1) = 4

In total, you have 8 ways to arrange Michael and David

Step 1:  Determine the probability = # of Desired Outcomes / # of Total Outcomes = 8 / 20 = 40%

Ratikainen · Answer

To pick David and Michael out of 6 people we have 6C2=15 
To arrange David and Michael in different group we have 3C1*3C1=9
Therefore (15-9)/15=40%

bumpbot · Answer

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A committee that includes 6 members is about to be divided into 2 subc

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