Question

(a) How many ways can a parent distribute five one-dollar bills to her three children?

(b) How many ways can she accomplish this if each child gets at least one dollar?

The answers are 21 and 6 I just need the work

Answer 1

(a) Let us see in how many ways the 5 dollars can be distributed among three children.

(i) 5 + 0 + 0 (1st gets all 5 , 2nd gets nothing and the 3rd gets nothing). This can be done in another 2 ways i.e (050) and (005). Therefore for this distribution there are 3 ways of distributing.

(ii) 4 + 1 + 0 ((1st gets 4 dollars , 2nd gets a dollar and the 3rd gets nothing). This can be done in another 5 ways i.e (401) (104) (140) (014) (041). Therefore for this distribution there are 6 ways of distributing.

(iii) 3 + 2 + 0 ((1st gets 3 dollars , 2nd gets 2 dollars and the 3rd gets nothing). This can be done in another 5 ways i.e (302) (203) (230) (023) (032). Therefore for this distribution there are 6 ways of distributing.

(iv) 3 + 1 + 1 ((1st gets 3 dollars , 2nd gets a dollar and the 3rd gets a dollar). This can be done in another 2 ways i.e (131) (113). Therefore for this distribution there are 3 ways of distributing.

(v) 2 + 2 + 1 (1st gets 2 dollars , 2nd gets 2 dollars and the 3rd gets a dollar). This can be done in another 2 ways i.e (122) (121). Therefore for this distribution there are 3 ways of distributing.

Therefore Total number of ways = 3 + 6 + 6 + 3 + 3 = 21 ways

(The other method is by partitioning where we see that the 5 dollars D D D D D, need at lest 2 partitions for 3 children D D D D D. So basically there are 5 Dollars and 2 partitions and therefore the number of ways = 7! / (5! * 2!) = 21)

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(b) Since each child has to get 1 dollar, let us assume that we have given each child the required dollar. Now we have 2 dollars left, which needs to be distributed.

(i) 2 + 0 + 0 (1st gets 2 dollars , 2nd gets 0 dollars and the 3rd gets 0 dollars). This can be done in another 2 ways i.e (020) (002). Therefore for this distribution there are 3 ways of distributing.

(ii) 1 + 1 + 0 (1st gets 1 dollar , 2nd gets 1 dollar and the 3rd gets 0 dollars). This can be done in another 2 ways i.e (101) (011). Therefore for this distribution there are 3 ways of distributing.

Therefore the total number of ways = 3 + 3 = 6 ways

(Using the rule of partition where we again need 2 partitions for 3 children, we have 2 dollars and 2 partitions. Therefore the total number of ways = 4! / (2! * 2!) = 6)

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In (a) and (b), we are dividing by a value which is equal to the value of number of dollars (because all dollars are similar) and by the value of partitions (because the partitions are similar).

This comes from the rule of permutations, which states that for an n letter word, which has a similar letters of one type, b similar letters of another type and so on, the total number of arrangements = n! / (a! * b! * ...)

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