Will of Honey

ways of distribution: [3 full barrels + 1 half barrel + 3 empty barrels]
[2 full barrels + 3 half barrels + 2 empty barrels]
[1 full barrel + 5 half barrels + 1 empty barrel] which adds up to 3 ways. now to get all that "distinct" ways all is left to do is 3!=3 * 2 * 1 = 6 ways.

there are 7 barrels full of honey, 7 barrels 1/2 full of honey and 7 empty barrels. It means that the total of the honey are (1 7 + 1/2 7 + 0*7) barrels, and it is equal to 10.5 barrels. Since we are going to use all barrels for 3 sons and they got the same amount, that means each son will get (10.5)/2 barrels, that is 3.5 barrels. These are the possibility for this case:

(A) (3 full + 1 half + 3 empty), (2 full + 3 half + 2 empty) and (2 full + 3 half + 2 empty) there are three sons to give the barrel, and a case where there are 2 same configurations (the 2 full + 3 half + 2 empty) so the ways sons can get that is (3!/2!) = 3 ways

(B) (1 full + 5 half + 2 empty), (3 full + 1 half +3 empty) and (3 full + 1 half +3 empty) here, we met the same case with the possibility (A), where there are 2 same configurations (the 3 full + 1 half + 3 empty) so the ways sons can get that is also (3!/2!) ways = 3 ways

Then, we add the result of possibility (A) and possibility (B). There, we got 3 ways + 3 ways = 6 ways

average 3.5 every son gets. It is 3f,1h 3e;3f 1h 3e; 1f 5h 1e; combination.it can be done in 3 ways or 2f 3h 2e; 2f 3h 2e ;3f 1h 3e combination. it can also be done in 3 ways . Total 6 ways here f is full ;h is half full & e is empty barrels.

Total amount = 7 full + 7 half + 7 empty = 10.5 with 21 barrels. Every person (all of them is 3 person) will have 3.5 with 7 barrels. Now the combination can be any order, one of them: 1. 3 full 1 half 3 empty, makes the remaining are 4 full 6 half 4 empty, then 2. 2 full 3 half 2 empty and 3. 2 full 3 half 2 empty. Another combination is 1. 3 full 1 half 3 empty, 2. 3 full 1 half 3 empty, and 3. 1 full 5 half 1 empty. I only can find these combination. Since each above combination can be arranged in two other arrangement, then 3+3= 6

let m be the number of full barrels, n be the number of 1/2 full barels, and p is the number of empty barrels. The total of amount of honey is 21/2. So each of son must get 7/2. And we have 2m+n+p = 7/2, and gives solution of triple (m,n,p) are (3,1,3), (1,5,1), (2,3,1). Then there are 2* ((3!)/(2!)(1!)) = 6 distinct ways such that the total barrel which received by 3 sons is 7 barrels.

we just take 3! ...to divide honey among 3 children

Total amount = 7 full + 7 half + 7 empty = 10.5 with 21 barrels. Every person (all of them is 3 person) will have 3.5 with 7 barrels. Now the combination can be any order, one of them: 1. 3 full 1 half 3 empty, makes the remaining are 4 full 6 half 4 empty, then 2. 2 full 3 half 2 empty and 3. 2 full 3 half 2 empty. Another combination is 1. 3 full 1 half 3 empty, 2. 3 full 1 half 3 empty, and 3. 1 full 5 half 1 empty. I only can find these combination. Since each above combination can be arranged in two other arrangement, then 3+3= 6

Solution is very simple just divide 7!/3! as 7 barrels are there and 3 sons......

A] 2(3 * 1+1 * .5+4 * 0) + 1 * 1+5 * .5+1 * 0 however last combination can be to any of the three......................................3
B] 2(2 * 1 +3 * .5 +2 * 0) +3 * 1+ * .5+3 * 0 however last combination can be to any of the three......................................3 ..................................................................................................... Total 6.