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If a coin is tossed six times, in how many ways can it fall four heads and two tails?
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
So far we have solved the basic combination problem of $r$ objects chosen from $n$ different objects. Now we will consider certain variations of this problem.
How many five-people committees consisting of 2 men and 3 women can be chosen from a total of 4 men and 4 women?
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:
2-Man Committees | 3-Woman Committees |
${M}_{1}{M}_{2}$ | ${W}_{1}{W}_{2}{W}_{3}$ |
${M}_{1}{M}_{3}$ | ${W}_{1}{W}_{2}{W}_{4}$ |
${M}_{1}{M}_{4}$ | ${W}_{1}{W}_{3}{W}_{4}$ |
${M}_{2}{M}_{3}$ | ${W}_{2}{W}_{3}{W}_{4}$ |
${M}_{2}{M}_{4}$ | |
${M}_{3}{M}_{4}$ |
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.
A high school club consists of 4 freshmen, 5 sophomores, 5 juniors, and 6 seniors. How many ways can a committee of 4 people be chosen that includes
Applying the multiplication axiom to the combinations involved, we get
We are choosing all 4 members from the 5 juniors, and none from the others.
$\mathrm{4C2}\cdot \mathrm{6C2}=\text{90}$
Since we don't want any freshmen on the committee, we need to choose all members from the remaining 16. That is
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.
How many five-letter word sequences consisting of 2 vowels and 3 consonants can be formed from the letters of the word INTRODUCE?
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 $5!$ .
$\mathrm{4C2}\cdot \mathrm{5C3}\cdot 5!=\text{7200}$ .
A standard deck of playing cards has 52 cards consisting of 4 suits each with 13 cards. In how many different ways can a 5-card hand consisting of four cards of one suit and one of another be drawn?
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
Ways of selecting a suit | Ways if selecting 4 cards from this suit | Ways if selecting the next suit | Ways of selecting a card from that suit |
$\mathrm{4C1}$ | $\text{13}\mathrm{C4}$ | $\mathrm{3C1}$ | $\text{13}\mathrm{C1}$ |
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