Correct Option (A)

To determine the number of zeroes at the end of a product, it is necessary to count the total number of prime factors 2 and 5 present in its prime factorization. Each pair of (2 × 5) contributes one zero. The number of zeroes will be equal to the minimum count between the total factors of 2 and the total factors of 5.

The given product is: $1 \times 5 \times 10 \times 15 \times 20 \times 25 \times 30 \times 35 \times 40 \times 45 \times 50 \times 55 \times 60$

Let's list the prime factors of 2 and 5 for each number in the product:

5: $5^{1}$ (one 5)
10: $2^{1} \times 5^{1}$ (one 2, one 5)
15: $3 \times 5^{1}$ (one 5)
20: $2^{2} \times 5^{1}$ (two 2s, one 5)
25: $5^{2}$ (two 5s)
30: $2^{1} \times 3 \times 5^{1}$ (one 2, one 5)
35: $5^{1} \times 7$ (one 5)
40: $2^{3} \times 5^{1}$ (three 2s, one 5)
45: $3^{2} \times 5^{1}$ (one 5)
50: $2^{1} \times 5^{2}$ (one 2, two 5s)
55: $5^{1} \times 11$ (one 5)
60: $2^{2} \times 3 \times 5^{1}$ (two 2s, one 5)

Total count of prime factor 5:

$1 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 1 + 2 + 1 + 1 = 14$

Total count of prime factor 2:

$1 + 2 + 1 + 3 + 1 + 2 = 10$

The number of zeroes at the end of the product is the minimum of the total counts of prime factors 2 and 5.

Minimum (10, 14) = 10.

Therefore, there are 10 zeroes at the end of the given product.

Incorrect Options:

Option B (12): This count would be obtained if there were 12 pairs of (2 × 5) factors. This could happen if either the count of factors of 2 or 5 was incorrectly determined to be 12, or if the minimum of the two counts was erroneously calculated as 12. For instance, if one only counted the factors of 5 from numbers ending in 5 or 0, but missed some factors of 2, or miscounted the factors of 5 from 25 and 50.
Option C (14): This value corresponds to the total number of prime factors 5 present in the product. However, the number of zeroes is limited by the lesser count of either prime factor 2 or 5. Since there are only 10 factors of 2, 14 zeroes cannot be formed.
Option D (15): This count is incorrect as it overestimates the number of pairs of (2 × 5) factors. A calculation leading to 15 zeroes would imply a significant miscount of either the factors of 2 or 5, or both, exceeding their actual minimum availability.