# In how many ways can $7$ different objects be divided among three persons so that either one or two of them do not get any objects?

In how many ways can $7$ different objects be divided among three persons so that either one or two of them do not get any objects?

My Approach

I am not able to understand how to solve the problem.

I did $7$!/$3$! . $4$!

Can anyone guide me how to solve the problem?

Call the people A, B, and C. Suppose C is to be shut out. Then A can get any of the $2^7$ subsets of the set of objects.

Our first estimate of the number of ways is therefore $3\cdot 2^7$. However, this sum counts $6$ times the number of ways two of the people receive nothing. However, there are only $3$ such ways, so the total is actually $3\cdot 2^7 -3$.

Another way: There are $2^7-2$ ways that C is shut out of the game and A and B each receive at least one object. Multiply by $3$, and add the three ways in which one of the players gets everything.

Remark: Our first way used the principle of Inclusion/Exclusion, because it is such a useful idea. The second way, for this problem, is more straightforward.

• Nicolas Can you explain me in more detail? – justin takro Nov 14 '15 at 2:57
• For the second way, we want to count the number of ways C gets nothing, and A and B each get at least one thing. Our collection of $7$ different objects has $2^7$ subsets. A is allowed to have (almost) any subset, with B getting the rest, The $2$ subsets A cannot get are (i) the empty set, for then A would get nothing and (ii) the full set, for then B would get nothing. So the number of possible choices for what A gets is $2^7-2$. Multiply by $3$, since any of the people can be shut out, and add $3$ for the ways one person gets all the toys. Total $3(2^7-2)+3=3\cdot 2^7-3$. – André Nicolas Nov 14 '15 at 3:06
• That is exactly what the answer solves. Look at second approach. We saw that there are $2^7-2$ ways in which C gets nothing, and each of A or B get at least one thing. There are also $2^7-2$ ways in which B gets nothing but A and C get something, and there are also $2^2-2$ ways in which A gets shut out and the others don't. Total so far, $3(2^7-2)$. Now we must count also the $3$ ways two people get shut out. So total $3(2^7-2)+3=381$. – André Nicolas Nov 14 '15 at 4:06
• There are $7$ distinct objects. How many ways are there to divide them between X and Y? We pick a subset of the set of objects, give it to X, giv the rest to Y. A $7$-element set has $2^7$ subsets. Except that in Way 2 we did not allow either X or Y to get nothing, which leaves $2^7-2$ ways. – André Nicolas Nov 14 '15 at 4:11
• Imagine the $7$ toys are lined up in a row, and we are deciding which ones to give to X. In front of each toy we write 1 if the toy is to go to X, and 0 if it isn't. There are $2^7$ different strings of 0's and 1's that we could write, and each string corresponds to X getting certain of the toys. – André Nicolas Nov 14 '15 at 4:16