Larry, his twin sisters, and 7 other people are to be divided into two

Question

Larry, his twin sisters, and 7 other people are to be divided into two teams of five people each. If Larry’s twin sisters cannot be assigned to opposite teams, how many different groups of 4 people could be on Larry’s team?

(A) 21
(B) 35
(C) 56
(D) 70
(E) 126

(A) 21
(B) 35
(C) 56
(D) 70
(E) 126

nemoexplicabo · Answer

Hi there!

This is a great combinations problem that tests your ability to handle constraints. The key concept here is restricted combinations — when certain elements have placement rules, we need to account for them systematically.

Let me walk through this step-by-step:

Step 1: Understand the setup
We have 10 people total: Larry, his 2 twin sisters, and 7 others. We're forming Larry's team of 5 people, which means Larry is already on the team. We need to choose 4 more people from the remaining 9.

Step 2: Identify the constraint
The twin sisters cannot be on opposite teams. This means:
- Both sisters are with Larry, OR
- Both sisters are on the other team, OR 
- One sister is with Larry and one sister sits out... wait, no — everyone must be on a team!

So really: both sisters together with Larry, or both sisters together on the opposite team.

Step 3: Use complementary counting (the easier approach)
Instead of counting valid arrangements directly, let's count total arrangements and subtract invalid ones.

Total ways to choose 4 people from the 9 remaining (ignoring constraints) = C(9,4) = 126

Step 4: Subtract invalid arrangements
Invalid = twin sisters on opposite teams. This means exactly one sister is with Larry.
- Choose 1 sister from 2: C(2,1) = 2
- Choose 3 more people from the remaining 7: C(7,3) = 35
- Invalid arrangements = 2 × 35 = 70

Step 5: Calculate the answer
Valid arrangements = 126 - 70 = 56

Answer: (C) 56

Common trap: Many students try to add the cases (both sisters with Larry + both sisters away) instead of using complementary counting, which leads to calculation errors. The constraint "cannot be on opposite teams" is actually easier to handle by identifying what violates it.

Takeaway: When you see "cannot be separated" or "cannot be on opposite sides" in combinations problems, complementary counting (Total - Violations) is often cleaner than case-by-case addition.

Hope this helps! Let me know if you have questions.

Ujjwalk31 · Answer

Larry is fixed on one team of 5, needing 4 more from 9 people: his 2 identical twin sisters (S1, S2) and 7 others. 
The sisters cannot be split across teams, so either both join Larry's team or both go to the other team.

Valid Cases

Both sisters with Larry: Choose 2 more from 7 others,
 7C2  = 21 ways.

Neither sister with Larry: Choose 4 from 7 others, 
 7C4  = 35 ways.

Total:

Add cases: 21+35=56 ways.

ANS = C

poojaarora1818 · Answer

Given: Larry, twin sisters and 7 other people are be divided into the team of two of five people each.

Constrain: Twin sisters cannot be assigned to the opposite teams.

To Find: Different groups of 4 people on Larry's team.

Case 1 (a) Larry and his twin sisters are already placed in the group.
 
(b) We have to select two people from the group of 7 => 7C2 = 21.

Case 2. (a) Larry is already placed without his twin sisters.

(b) We have to select three people from the group of seven => 7C3 = 35.

Now, adding case 1 & case 2 we get 56.

Answer: Option C

nemoexplicabo · Answer

This is a combinatorics problem dealing with identical individuals and unlabeled groups.

Key concept being tested: counting with identical objects in combinations.

1. There are 10 people: Larry, his twin sisters (indistinguishable), and 7 others. The problem asks for the number of ways to divide them into two teams of five, recognizing that the teams are unlabeled (Team A vs Team B doesn’t matter). 
2. First treat everyone as distinct. The number of ways to choose 5 people out of 10 to form one team is C(10,5) = 252. Because the two teams are unlabeled, each division is counted twice (Team X vs Team Y), so divide by 2 to get 126. 
3. Now adjust for the identical twins. In the count above, swapping the twins between positions doesn’t create a new arrangement, so each valid division is overcounted by a factor of 2. Divide again by 2 to get 63. 
4. Alternatively, break into cases: either both twins are on Larry’s team or both are on the other team. If both twins join Larry, choose 2 more from the 7 others: C(7,2) = 21. If neither twin joins Larry, choose 4 from 7: C(7,4) = 35. Sum gives 56. Since the question asks for groups of 4 joining Larry (Larry fixed), this matches the case method answer.

Common trap: Many students forget to divide by 2 for unlabeled teams or to account for the twins being identical. This leads to double-counting.

Takeaway: In grouping problems, always consider whether the groups are distinct and whether any individuals are identical. Adjust your count accordingly to avoid overcounting.

Larry, his twin sisters, and 7 other people are to be divided into two

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Larry, his twin sisters, and 7 other people are to be divided into two

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