Give a combinatorial (logical) justification for this assertion:
The right hand side tells us that we have to select persons out of a group of persons. To do so, we consider any particular group of persons from these persons. Specify these persons by the symbols .
Now, to count all the possible r-groups from this group of , we consider the following mutually exclusive cases:
(1) The -group does not contain
Such r-groups can be formed in ways since we have to select people out of .
(2) The -group contains but not
We have to select people out of because we already have selected so we need only more people and since we are not taking , we have people to choose from. This can be done ways.
(3) The -group contains , but not
We now have to select people out of . This can happen in ways. Proceeding in this way, we arrive at the last two possible cases.
(r) The -group contains but not .
We need to select only person out of available for selection. This can be done in ways.
(r+1) The -group contains
In this case, our -group is already complete. We need not select any more person. This can be done in or equivalently way.
Convince yourself that these cases cover all the possible cases that can arise in the formation of the – groups. Also, all these cases are mutually exclusive. Thus, adding the number of possibilities of each case will give us the total number of -groups possible, i.e.
How many distinct throws are possible with a throw of dice which are identical to each other, i.e. indistinguishable among themselves ?
The important point to be realised here is that the dice are totally identical. Suppose we had just dice, say die and die . Suppose that, upon throwing these dice, we get a “two” on and a “three” on . This case would be the same as the one where we get a “three” on and a “two” on because we cannot distinguish between and . What we are concerned with is only what numbers show up on the top of the dice. We are not concerned with which die shows what number. This means that if we have dice and we throw them, we are only concerned with how many “ones”, “twos”, “threes” etc show on the top faces of the dice; we are not at all interested in which die throws up what number.
If we denote the number of “ones” we get by , number of “twos” we get by and so on, we will have
Thus, the total number of distinct throws will be simply the number of non-negative solutions to this integral equation.
As discussed earlier, this number will be .
What would be the number of distinct throws if the dice were not identical?
Give combinatorial arguments to prove that
Let us first interpret what the left hand side of this assertion says.
Suppose we have a group of people. We select a sub-group of size from the group of people. This can be done in ways. Once the sub-group has been formed, we select a leader of that sub-group, and send that sub-group on an excursion. The leader can be selected in ways. Thus, the total number of different ways in which an – group can be formed with a unique leader is .
Now can take any integer value from to , i.e. . Thus, the total number of all possible sub-groups, each sub-group being assigned a unique leader, will be which is the left hand side of our assertion.
To prove this equal to the right hand side, we count the sub-groups from a different angle. We count all those sub-groups in which a particular person, say , is the leader.
Since is the leader, is fixed in our sub-group. For each of the remaining people, we have two options. We either put the person in the group led by or we don’t. Thus, the total number of sub-groups in which is the leader will be
Since any of the persons can be the leader, and under each person’s leadership, groups can be formed, the total number of sub-groups, each sub-groups under some unique person’s leadership is . This proves our assertion that