How many partitions of $12$ are there that have at least four parts, such that the largest, second-largest, third-largest, and fourth-largest parts are respectively greater than or equal to $4,3,2,1$?

4 , 4 , 2 , 2

4, 3 , 3 , 2

Step-by-step explanation:

Given data in the problem:

total value = 12

now,

each partition should be greater than or equal to 4 , 3 , 2 , 1

now, the sum of the above terms comes as:

4 + 3 + 2 + 1 = 10

difference from the original values : 12 - 10 = 2

so we can divide 2 into two equal parts to the number 3 and 1

thus, we get the partitions as:

4 , (3 + 1) , 2 , (1 + 1) = 4 , 4 , 2 , 2

and another partition can be

4, 3, (2 + 1), (1 + 1) = 4, 3 , 3 , 2

(a) the volume of the cone is 20/3 in3

hence, option a is correct.

2nd one in the first row is correct