Question

Which of the following statements individually provide enough information to determine the number of students in a group?
Indicate all such statements.
[Note: Select one or more answer choices]

• A. The number of ways 3 students can be selected from the group to form a team is 35.
• B. The number of ways 3 students from the group can be seated in a row is 210.
• C. The number of ways all the students from the group can be selected to form a team is 1.
• D. The number of ways 3 students can be selected from the group to form a team equals the number of ways 4 students can be selected from the group to form a team.
• E.

Hint:

Remember the key difference between combinations (team selection, committee formation) and permutations (arrangements, seating, assigning specific roles). Also, memorize the useful combinatorial identity: \(\binom{n}{r} = \binom{n}{n-r}\). The property used in statement D, \(\binom{n}{r} = \binom{n}{k} \implies n = r+k\) (for \(r \neq k\)), is a direct consequence of this identity.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question tests our understanding of combinations (selections, order doesn't matter) and permutations (arrangements, order matters). We need to analyze each statement to see if it leads to a unique value for \(n\), the total number of students in the group.
Step 2: Key Formula or Approach:
Let \(n\) be the total number of students.
- The number of ways to select \(r\) items from \(n\) (combination) is given by \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).
- The number of ways to arrange \(r\) items from \(n\) (permutation) is given by \( P(n,r) = \frac{n!}{(n-r)!} \).
We will formulate an equation for \(n\) based on each statement and check if it can be solved for a unique, positive integer value of \(n\).
Step 3: Detailed Explanation:
Statement A: The number of ways 3 students can be selected from the group to form a team is 35.
"Selection" implies combination. So, we have \(\binom{n}{3} = 35\).
\[ \frac{n(n-1)(n-2)}{3 \cdot 2 \cdot 1} = 35 \] \[ n(n-1)(n-2) = 35 \times 6 = 210 \] We need to find three consecutive integers whose product is 210. We can estimate: \(6\^{}3 = 216\), so the numbers should be around 6. Let's try \(n=7\): \(7 \times 6 \times 5 = 210\). This works. Since the function \(f(n) = n(n-1)(n-2)\) is strictly increasing for \(n>2\), there is only one positive integer solution for \(n\). Thus, \(n=7\). This statement is sufficient.
Statement B: The number of ways 3 students from the group can be seated in a row is 210.
"Seated in a row" implies arrangement, which is a permutation. So, we have \(P(n,3) = 210\).
\[ \frac{n!}{(n-3)!} = 210 \] \[ n(n-1)(n-2) = 210 \] This is the same equation as in Statement A. We already found that the unique positive integer solution is \(n=7\). This statement is sufficient.
Statement C: The number of ways all the students from the group can be selected to form a team is 1.
This means selecting \(n\) students from a group of \(n\). The number of ways is \(\binom{n}{n}\).
\[ \binom{n}{n} = \frac{n!}{n!(n-n)!} = \frac{n!}{n!0!} = 1 \] (since \(0!=1\)). This equation is true for any positive integer \(n\). It does not allow us to determine a unique value for \(n\). This statement is not sufficient.
Statement D: The number of ways 3 students can be selected from the group to form a team equals the number of ways 4 students can be selected from the group to form a team.
This gives the equation \(\binom{n}{3} = \binom{n}{4}\).
Using the property that \(\binom{n}{r} = \binom{n}{k}\) implies either \(r=k\) (which is not the case here) or \(n = r+k\).
\[ n = 3 + 4 = 7 \] This gives a unique value for \(n\). This statement is sufficient.
Step 4: Final Answer:
Statements A, B, and D each provide enough information to uniquely determine the number of students. Statement C does not.