Question

The number of ways of arranging 3 red, 2 white, and 4 blue flowers of different sizes into a garland such that no two blue flowers come together is:

• A. \(6! \times 36\)
• B. \(5! \times 24\)
• C. \(7! \times 70\)
• D. \(9! \times 84\)

Hint:

For circular arrangements, use \((n-1)!\) for \(n\) distinct objects. When no two objects of the same kind can be adjacent, place the other objects first and then insert the restricted ones in the gaps.

Answer 1

The Correct Option is A

To solve the problem of arranging 3 red, 2 white, and 4 blue flowers such that no two blue flowers come together, follow these steps:

1. Arrange non-blue flowers first:
   The non-blue flowers are 3 red and 2 white flowers. Total non-blue flowers = 3 + 2 = 5.
2. Number of ways to arrange non-blue flowers:
   The 5 flowers can be arranged in \(5!\) ways.
3. Create gaps for blue flowers:
   Arranging the 5 non-blue flowers creates 6 possible gaps (before, between, and after the flowers) for placing blue flowers.
4. Choose gaps for blue flowers:
   We need to place 4 blue flowers such that no two are adjacent. Thus, we select 4 gaps from the 6 available options. This can be done in \(\binom{6}{4}\) ways.
5. Arrange blue flowers within chosen gaps:
   The 4 blue flowers can be arranged in the chosen gaps in \(4!\) ways.
6. Total arrangements:
   By multiplying these results, the total number of arrangements is:
   \(5! \times \binom{6}{4} \times 4! = 120 \times 15 \times 24 = 6! \times 36\).

The number of ways to arrange the flowers such that no two blue flowers are together is \(6! \times 36\).