0.10 Sets and counting  (Page 9/9)

First we solve this problem using [link] technique–permutations with similar elements.

We need 4 heads and 2 tails, that is

There are $\frac{6!}{4!2!}=\text{15}$ permutations.

Now we solve this problem using combinations.

Suppose we have six spots to put the coins on. If we choose any four spots for heads, the other two will automatically be tails. So the problem is simply

$\mathrm{6C4}=\text{15}$ .

Incidentally, we could have easily chosen the two tails, instead. In that case, we would have gotten

$\mathrm{6C2}=\text{15}$ .

Further observe that by definition

and $\mathrm{6C2}=\frac{6!}{4!2!}$

Which implies

We list 4 men and 4 women as follows:

Since we want 5-people committees consisting of 2 men and 3 women, we'll first form all possible two-man committees and all possible three-woman committees. Clearly there are 4C2 = 6 two-man committees, and 4C3 = 4 three-woman committees, we list them as follows:

For every 2-man committee there are four 3-woman committees that can be chosen to make a 5-person committee. If we choose $M_1{M}_2$ as our 2-man committee, then we can choose any of $W_1{W}_2{W}_3$ , $W_1{W}_2{W}_4$ , $W_1{W}_3{W}_4$ , or $W_2{W}_3{W}_4$ as our 3-woman committees. As a result, we get

Similarly, if we choose $M_1{M}_3$ as our 2-man committee, then, again, we can choose any of $W_1{W}_2{W}_3$ , $W_1{W}_2{W}_4$ , $W_1{W}_3{W}_4$ , or $W_2{W}_3{W}_4$ as our 3-woman committees.

And so on.

Since there are six 2-man committees, and for every 2-man committee there are four 3- woman committees, there are altogether $6\cdot 4=24$ five-people committees.

In essence, we are applying the multiplication axiom to the different combinations.

1. Applying the multiplication axiom to the combinations involved, we get

   $\mathrm{4C1}\cdot \mathrm{5C1}\cdot \mathrm{5C1}\cdot \mathrm{6C1}=\text{600}$

2. We are choosing all 4 members from the 5 juniors, and none from the others.

   $\mathrm{5C4}=5$

3. $\mathrm{4C2}\cdot \mathrm{6C2}=\text{90}$

4. Since we don't want any freshmen on the committee, we need to choose all members from the remaining 16. That is

   $\text{16}\mathrm{C4}=\text{1820}$

5. Of the 4 people on the committee, we want at least three seniors. This can be done in two ways. We could have three seniors, and one non-senior, or all four seniors.

   $\mathrm{6C3}\cdot \text{14}\mathrm{C1}+\mathrm{6C4}=\text{295}$

First we select a group of five letters consisting of 2 vowels and 3 consonants. Since there are 4 vowels and 5 consonants, we have

Since our next task is to make word sequences out of these letters, we multiply these by $5!$ .

$\mathrm{4C2}\cdot \mathrm{5C3}\cdot 5!=\text{7200}$ .

We will do the problem using the following steps. Step 1. Select a suit. Step 2. Select four cards from this suit. Step 3. Select another suit. Step 4. Select a card from that suit.

Applying the multiplication axiom, we have