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Combination

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☼ COMBINATION  :   When  we have to choose a certain number of  objects from a given set of  objects in such a way that the order of  selection is not important, the selection is known as a combination. Consider the following example.

 e.g. There were three persons in a meeting. Each person shook hands with every other person. Find the total number of  ways this happened.

 Soln. Let, the persons were A,  B  and  C.  They  shook hand in the way : A \large \leftrightarrow B, A \large \leftrightarrow C, B \large \leftrightarrow C

i.e.   Only  three ways are possible.

\large \mapsto In  the above example, 2 persons were needed from a group of  3 persons for shake hand  such a way that the order of  the arrangements is not important.

( i.e. A \large \leftrightarrow B  &  B \large \leftrightarrow A are same. )

Now, if there were  n = 3 persons and  r  = 2 persons were to be selected, then we could make the selection in  \frac{3 \times 2}{2} =   3  ways.

This can also be written as    \frac{3 \times 2 \times 1}{\left (1 \right ) \times \left (2\times 1 \right )}    =     \frac{3!}{1! \times 2!}      =       \frac{3!}{\left (3-2 \right )! \times 2!}

\therefore Number of  combinations   = _{n}C_{r}   =  \frac{n!}{\left (n-r \right )! \times r!}

■    Number of  permutations  =  Number of  combinations \times  r !  (  i.e.    _{n}C_{r}  =  _{n}P_{r}  \times r !  )

■    _{n}C_{r}   =  \frac{_{n}P_{r}}{r!}   =  \frac{n!}{\left (n-r \right )! \times r!}     ;   0  \leq  r  \leq  n

■  _{8}C_{3}    = \frac{8\times 7 \times 6}{3!} =  56,  _{12}C_{5}  = \frac{12 \times 11 \times 10 \times 9 \times 8}{5!} = 792, _{5}C_{1}  = 5, _{9}C_{0}  = 1 = _{9}C_{9}

■     The distinct symbols for combination   =   _{n}C_{r}   =   ^{n}C_{r}    =  \left (\frac{n}{r} \right )

■    We can find the distinct value of    by Pascal triangle also

■  _{n}C_{r}    =  _{n}C_{n-r}   &   _{n}C_{r}   +  _{n}C_{r+1}    =    _{n+1}C_{r+1}

 e.g. (i) In a window there are m vertical and n horizontal bars. Find the number of  rectangles formed. (m, n  2)

  Soln. By selecting 2 vertical bars and 2 horizontal bars a rectangle can be formed.

\therefore The required no. of  rectangles  are = _{m}C_{2} \times _{n}C_{2}

= \frac{m\times \left (m-1 \right )}{2!} \times \frac{n\times \left (n-1 \right )}{2!}

= \frac{mn(m-1)(n-1)}{4}

  (ii) Find the number of  diagonals in a polygon of   n  sides (or n vertices).                                                                                     (n  \geq  3 )

  Soln. Here, no. of  sides of  a polygon  =  n

&    Total no. of  line-segment joining any two points from given

n  points  = _{n}C_{2}

\therefore The required no. of  diagonals  are =  _{n}C_{2}   –  No. of  sides

=   \frac{n\times\left (n-1 \right )}{2!}-n

=  \frac{n}{2}  [(n – 1) – 2]

=   \frac{n(n-3)}{2}

  (iii) In how many ways can 3 mangoes and  2  apples be selected from  7  mangoes and  4  apples to form a basket of  5 fruits ?

  Soln. 3 mangoes can be selected from  7  mangoes in  _{7}C_{3}  ways and  2  apples can be selected from  4  apples in  _{4}C_{2}  ways.

Total number of  ways  = _{7}C_{3}   \times   _{4}C_{2} =   35    6   =   210

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