Question

in how many ways can 5 different coins be placed into three different boxes such that each box has at least one coin placed in it

Answer 1

First, the usual one of checking different arrangements with a 3,1,1 distribution and the other with a 2,2,1 distribution and calculating their respective combinations. Works out to a total of 108. Again a long method. So wanted a shorter one.

Then i tried the "Stars & Bars" approach. I first gave each row a ball so am left with 2 balls and 6 spaces to fill.

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Answer 2

Hey mate,

In how many ways can 5 distinct balls be distributed into 3 distinct boxes such that each box has atleast 1 ball in it?

Suppose the balls are labelled 1, 2, . . . , 5 and the boxes are labelled A, B, and C.

Since there are 5 balls and 3 boxes, and each box should have atleast 1 ball, we can form 4C2 (=6) different cases for the arrangement of the balls into the boxes :

For case 1 : Box A - 1 ball, Box B - 1 ball and Box C - 3 balls.

We have 5 balls, thus, one ball can be taken by box A in 5C1 ways. Out of the remaining 4 balls, one ball can be taken by box B in 4C1 ways and the remaining balls can be placed in the 3rd box in 3C3 (=1) way.

Similarly for 2nd case, the number of combinations = 5C1×4C2×2C2

For 3rd : the number of combinations = 5C1×4C3

For 4th: the number of combinations = 5C2×3C1

For 5th: the number of combinations = 5C2×3C2

For 6th: the number of combinations = 5C3×2C1

Hence, the total number of ways = 5C1{4C1+4C2+4C3}+5C2{3C1+3C2}+5C3{2C1}

= 70+60+20 = 150

We can also find the number of ways using the formula of Stirling numbers of the second kind, denoted by S(n,k) or {nk}.

The number of ways to distribute n distinct balls into k distinct boxes such that each box has atleast 1 ball = k!S(n,k)

Thus, the number of ways to distribute 5 distinct balls into 3 distinct boxes such that each box has atleast 1 ball = 3!S(5,3) = 6×25 = 150

S(5,3)=25 (from table of values of Stirling numbers of the second kind)

Since the number of balls and boxes aren't too many, instead of directly substituting the value, we can also solve S(5,3).

S(n,k)=1k!∑j=0k(−1)k−j(kj)(j)n

Thus, the number of ways to distribute n distinct balls into k distinct boxes such that each box has atleast 1 ball = k!1k!∑j=0k(−1)k−j(kj)(j)n

= ∑j=0k(−1)k−j(kj)(j)n

For (n = 5 and k = 3), the number of ways = ∑j=03(−1)3−j(3j)(j)5

= (−1)3(30)05+(−1)2(31)15+(−1)1(32)25+(−1)0(33)35

= (31)15−(32)25+(33)35

Where,

In (31)15, (31) represents the number of ways we can choose 1 box out of 3 distinct boxes and 15 represents the number of ways in which 5 balls can be distributed to that box.

In (32)25, (32) represents the number of ways we can choose 2 boxes out of 3 and 25 represents the number of ways in which 5 distinct balls can be distributed to those 2 distinct boxes with empty boxes allowed.

Similarly, 35 represents the number of ways in which 5 distinct balls can be distributed to 3 distinct boxes with empty boxes allowed.

Final result = 3−3(32)+243 = 150.

Hope it will help you.

✨It's. M.S.V.