Step 1: Represent the problem. We have 3 sets of twins $\{A,A\},\{B,B\},\{C,C\}$ and $2$ empty chairs. The twins in each pair must sit together, so each pair behaves like a "block." Thus we have: - 3 twin blocks ($AA, BB, CC$), - 2 empty chairs. So, total $5$ objects to arrange around a circular table.
Step 2: Circular arrangements. For $n$ distinct objects around a circle, the number of arrangements is $(n-1)!$. Here: \[ (5-1)! = 4! = 24. \]
Step 3: Adjust for indistinguishability within pairs. Within each twin pair, the order doesn't matter (since twins are indistinguishable). Thus, no further division is needed because each pair is already treated as a block. However, we must also note that the two empty chairs are indistinguishable. So we divide by $2!$: \[ \frac{24}{2} = 12. \]
Step 4: Final Answer. Therefore, the number of unique seating arrangements is:
\[ \boxed{12} \]
How many possible words can be created from the letters R, A, N, D (with repetition)?
Let R = {(1, 2), (2, 3), (3, 3)} be a relation defined on the set \( \{1, 2, 3, 4\} \). Then the minimum number of elements needed to be added in \( R \) so that \( R \) becomes an equivalence relation, is:}
The 12 musical notes are given as \( C, C^\#, D, D^\#, E, F, F^\#, G, G^\#, A, A^\#, B \). Frequency of each note is \( \sqrt[12]{2} \) times the frequency of the previous note. If the frequency of the note C is 130.8 Hz, then the ratio of frequencies of notes F# and C is:
Here are two analogous groups, Group-I and Group-II, that list words in their decreasing order of intensity. Identify the missing word in Group-II.
Abuse \( \rightarrow \) Insult \( \rightarrow \) Ridicule
__________ \( \rightarrow \) Praise \( \rightarrow \) Appreciate