The number of combinations of n objects taken r at a time is given by
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Chosing subsets its not about permutations and repeated permutations, but it's just about combinations. Your problem is quite easy to understand practically - answer: $3$ ($\{B, R\}, \{R, R\}, \{B, B\}$). Remember that (unordered) sets chosen don't count distinctly if you change order ($\{B, R\} = \{R, B\}$). Show To solve this, in general case, with $k$ type of objects, $m$ objects in a subset, $n_m$ objects with at least $m$ occurencies, $n_{m - 1}$ objects with at least $m - 1$ occurencies and so on may not be as easy as understanding a practical idea, as these require analytical solutions (there are combinatorial objects regarding multisets, not simple sets). Simple is to understand and solve the case with $\alpha$ types of objects with at least $2$ occurences and $\beta$ types of objects with $1$ occurence, chosen in . There are $\binom{\alpha}{2}$ ways to choose pairs of identical objects and $\binom{\alpha + \beta}{2}$ ways to choose pairs of distinct objects. So, our combinatorial object is defined as $\binom{\alpha, \beta}{2} = \binom{\alpha}{2} + \binom{\alpha + \beta}{2}$, a generalisation of standard combinations. Note that $\binom{n}{r}$ is the binomial coefficient, an expression of combinations. You should read more on these topics to understand the difference between different combinatorial objects like permutations, repetition permutations, combinations, arrangeaments (here order is important!) and nultiset combinatorics. \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)
Learning ObjectivesIn this section you will learn to 1. Count the number of combinations of \(\mathrm{r}\) out of \(\mathrm{n}\) items (selections without regard to arrangement ) 2. Use factorials to perform calculations involving combinations Suppose we have a set of three letters { A, B, C }, and we are asked to make two-letter word sequences. We have the following six permutations. AB BA BC CB AC CA Now suppose we have a group of three people { A, B, C } as Al, Bob, and Chris, respectively, and we are asked to form committees of two people each. This time we have only three committees, namely, AB BC AC When forming committees, the order is not important, because the committee that has Al and Bob is no different than the committee that has Bob and Al. As a result, we have only three committees and not six. Forming word sequences is an example of permutations, while forming committees is an example of combinations - the topic of this section. Permutations are those arrangements where order is important, while combinations are those arrangements where order is not significant. From now on, this is how we will tell permutations and combinations apart. In the above example, there were six permutations, but only three combinations. Just as the symbol nPr represents the number of permutations of n objects taken r at a time, nCr represents the number of combinations of n objects taken r at a time. So in the above example, 3P2 = 6, and 3C2 = 3. Our next goal is to determine the relationship between the number of combinations and the number of permutations in a given situation. In the above example, if we knew that there were three combinations, we could have found the number of permutations by multiplying this number by 2!. That is because each combination consists of two letters, and that makes 2! permutations. Example \(\PageIndex{1}\)Given the set of letters { A, B, C, D }. Write the number of combinations of three letters, and then from these combinations determine the number of permutations. Solution We have the following four combinations. ABC BCD CDA BDA Since every combination has three letters, there are 3! permutations for every combination. We list them below. \[\begin{array}{cccc} The number of permutations are 3! times the number of combinations; that is 4P3 = 3! \(\cdot\) 4C3 or \[4 \mathrm{C} 3=\frac{4 \mathrm{P} 3}{3 !} \nonumber \] In general, \[\mathrm{nCr}=\frac{\mathrm{nPr}}{\mathrm{r} !} \nonumber \] Since \[\mathrm{nPr}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !} \nonumber \] We have, \[\mathrm{nCr}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !} \nonumber \] Summarizing, Note1. Combinations A combination of a set of elements is an arrangement where each element is used once, and order is not important. 2. The Number of Combinations of n Objects Taken r at a Time \[\mathrm{nCr}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !} \nonumber \] where \(\mathrm{n}\) and \(\mathrm{r}\) are natural numbers. Example \(\PageIndex{3}\)Example \(\PageIndex{2}\)Compute:
Solution We use the above formula. \[5 \mathrm{C} 3=\frac{5 !}{(5-3) ! 3 !}=\frac{5 !}{2 ! 3 !}=10 \nonumber \] \[7 \mathrm{C} 3=\frac{7 !}{(7-3) ! 3 !}=\frac{7 !}{4 ! 3 !}=35 \nonumber \] Example \(\PageIndex{3}\)In how many different ways can a student select to answer five questions from a test that has seven questions, if the order of the selection is not important? Solution Since the order is not important, it is a combination problem, and the answer is 7C5 = 21 Example \(\PageIndex{4}\)How many line segments can be drawn by connecting any two of the six points that lie on the circumference of a circle? Solution Since the line that goes from point A to point B is same as the one that goes from B to A, this is a combination problem. It is a combination of 6 objects taken 2 at a time. Therefore, the answer is \[6 \mathrm{C} 2=\frac{6 !}{4 ! 2 !}=15 \nonumber \] Example \(\PageIndex{5}\)There are ten people at a party. If they all shake hands, how many hand-shakes are possible? Solution Note that between any two people there is only one hand shake. Therefore, we have 10C2 = 45 hand-shakes. Example \(\PageIndex{6}\)The shopping area of a town is in the shape of square that is 5 blocks by 5 blocks. How many different routes can a taxi driver take to go from one corner of the shopping area to the opposite cater-corner? Solution Let us suppose the taxi driver drives from the point A, the lower left hand corner, to the point B, the upper right hand corner as shown in the figure below.B ATo reach his destination, he has to travel ten blocks; five horizontal, and five vertical. So if out of the ten blocks he chooses any five horizontal, the other five will have to be the vertical blocks, and vice versa. Therefore, all he has to do is to choose 5 out of ten to be the horizontal blocks The answer is 10C5, or 252. Alternately, the problem can be solved by permutations with similar elements. The taxi driver's route consists of five horizontal and five vertical blocks. If we call a horizontal block H, and a vertical block a V, then one possible route may be as follows. HHHHHVVVVV Clearly there are \(\frac{10!}{5!5!}= 252\) permutations. Further note that by definition 10C5 = \(\frac{10!}{5! 5!}\). Example \(\PageIndex{7}\)If a coin is tossed six times, in how many ways can it fall four heads and two tails? Solution First we solve this problem using section 6.5 technique-permutations with similar elements. We need 4 heads and 2 tails, that is HHHHTT There are \(\frac{6!}{4!2!}= 15\) 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 6C4 = 15. Incidentally, we could have easily chosen the two tails, instead. In that case, we would have gotten 6C2 = 15. Further observe that by definition \[6 \mathrm{C} 4=\frac{6 !}{2 ! 4 !} \nonumber \] and \[6 \mathrm{C} 2=\frac{6 !}{4 ! 2 !} \nonumber \] Which implies 6C4 = 6C2. This page titled 4.4.3: Combinations is shared under a CC BY license and was authored, remixed, and/or curated by Rupinder Sekhon and Roberta Bloom. |
