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?
- (a) 18
- (b) 27
- (c) 54
- (d) 81 ✓ UPSC's answer
Why the answer is (d)
• The 9 cups are arranged in a 3x3 grid (3 rows and 3 columns).
• The condition is that each of the 3 rows must contain at least one cup of coffee.
• Since there are only 3 cups of tea, the maximum number of tea cups in any single row is 3. However, if a row had 3 tea cups, it would have 0 coffee cups, violating the condition.
• Therefore, the restriction is simply that no row consists entirely of tea cups.
• Total ways to arrange 6 identical coffee cups and 3 identical tea cups in 9 positions is $\binom{9}{3} = 84$.
• We subtract the cases where at least one row has no coffee (i.e., all 3 cups in that row are tea). There are 3 rows. If row 1 is all tea, the remaining 6 positions have 6 coffee cups, so 1 way. Similarly for row 2 and row 3. Since there are only 3 tea cups, two rows cannot be all tea simultaneously. Thus, invalid cases = 3.
• Valid arrangements = $84 - 3 = 81$.
Why the other options are wrong
- (a) 18
- 18 is incorrect because it does not account for the total permutations of the 3 tea cups among the 9 positions minus the invalid row cases.
- (b) 27
- 27 is incorrect as it likely results from an erroneous calculation of combinations or permutations without applying the row constraint correctly.
- (c) 54
- 54 is incorrect because it fails to correctly subtract the invalid configurations where a row contains only tea cups from the total combinations.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2022, held on 5 June 2022. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.