There are 9 cups placed on a table arranged in equal number of rows and columns out of which 6 cups contain coffee and 3 cups contain tea. In how many ways can they be arranged so that each row should contain at least one cup of coffee?

Correct Option (4)

The problem requires arranging 9 cups, consisting of 6 coffee cups and 3 tea cups, in a 3x3 grid. The constraint is that each row must contain at least one coffee cup.

A 3x3 grid implies 3 rows (R₁, R₂, R₃), with each row having 3 positions. The condition "each row should contain at least one cup of coffee" means that no row can be entirely filled with tea cups. Since there are only 3 tea cups in total, this implies that no row can contain all 3 tea cups.

We can determine the total number of valid arrangements by considering the possible distributions of the 3 tea cups across the three rows, ensuring that no row has 3 tea cups:

1. Case 1: Each row contains exactly one tea cup (1T, 1T, 1T).

   In this configuration, each row will have 1 tea cup and 2 coffee cups (since each row has 3 positions). This distribution inherently satisfies the condition of having at least one coffee cup per row.

   • For Row R₁: The number of ways to place 1 tea cup in 3 available positions is C(3,1) = 3 ways.
   • For Row R₂: The number of ways to place 1 tea cup in 3 available positions is C(3,1) = 3 ways.
   • For Row R₃: The number of ways to place 1 tea cup in 3 available positions is C(3,1) = 3 ways.

   The total number of arrangements for Case 1 = 3 × 3 × 3 = 27 ways.

2. Case 2: Two tea cups are in one row, one tea cup is in another row, and the third row has no tea cups (2T, 1T, 0T).

   This distribution also satisfies the condition for coffee cups:

   • The row with 2 tea cups will contain 1 coffee cup.
   • The row with 1 tea cup will contain 2 coffee cups.
   • The row with 0 tea cups will contain 3 coffee cups.

   To calculate the arrangements for this case:

   • First, determine which row receives 2 tea cups: C(3,1) = 3 ways.
   • Next, determine which of the remaining two rows receives 1 tea cup: C(2,1) = 2 ways.
   • The last remaining row automatically receives 0 tea cups: C(1,1) = 1 way.
   • This results in 3 × 2 × 1 = 6 distinct permutations for assigning the (2T, 1T, 0T) distribution to the rows.

   For each such permutation (e.g., Row R₁ gets 2T, Row R₂ gets 1T, Row R₃ gets 0T):

   • Number of ways to place 2 tea cups in a row of 3 positions: C(3,2) = 3 ways.
   • Number of ways to place 1 tea cup in a row of 3 positions: C(3,1) = 3 ways.
   • Number of ways to place 0 tea cups in a row of 3 positions: C(3,0) = 1 way.

   Thus, for one specific permutation of (2T, 1T, 0T) across the rows, there are 3 × 3 × 1 = 9 ways.

   The total number of arrangements for Case 2 = (Number of permutations of (2T, 1T, 0T) across rows) × (Ways to place cups within rows for one permutation)

   = 6 × 9 = 54 ways.

The only other possible distribution of 3 tea cups is (3T, 0T, 0T), where one row has all 3 tea cups. This scenario is invalid as it violates the condition that each row must contain at least one coffee cup.

Therefore, the total number of valid arrangements is the sum of arrangements from Case 1 and Case 2:

Total ways = 27 (from Case 1) + 54 (from Case 2) = 81 ways.

Incorrect Options:

The options 18, 27, and 54 represent incomplete calculations or specific sub-cases of the total valid arrangements. For instance, 27 accounts only for arrangements where each row contains exactly one tea cup. 54 accounts only for arrangements where the tea cups are distributed as (2, 1, 0) across the rows. 18 represents a partial sum within the (2,1,0) distribution, corresponding to arrangements where a specific row (e.g., R₁) has two tea cups, and the remaining two rows distribute the single remaining tea cup and zero tea cups. These individual values do not encompass all possible valid arrangements under the given conditions.