5.0 Combinations

  Permutations and Combinations

• Without repetition: Here the objects are unique.
• With repetition: Here the objects need not be unique.

Question 15. There are $12$ shirts - pant - stockings each of distinct color.

(A) In how many ways can you select a set of shirt, pant and stockings if mix and match is allowed.

(B) In how many ways can you select a set of shirt, pant and stockings if mix and match is allowed and if shirt pant and stockings should be each of different color?

(C) What is the maximum number of selections that can be made if in a set, shirt and pant be of the same color and stocking of different color?

Solution:

(A) One can choose one out of $12$ from shirt, one out of $12$ from pant and one out of $12$ from stockings to form a set.

Thus total selections possible is, $$\begin{equation} \begin{aligned} = {\;^{12}}{C_1}{ \times ^{12}}{C_1}{ \times ^{12}}{C_1} \\ = \frac{{12!}}{{1!\left( {12 - 1} \right)!}} \times \frac{{12!}}{{1!\left( {12 - 1} \right)!}} \times \frac{{12!}}{{1!\left( {12 - 1} \right)!}} \\ = {12^3} \\ = 1728 \\\end{aligned} \end{equation} $$

(B) One can select one shirt from $12$ distinct colors. This color cannot be chosen in pant.

Hence one out of $11$ pants can be chosen.

Now since two colors are eliminated, one out of $10$ stockings can be chosen.

Thus total selections is, $$\begin{equation} \begin{aligned} = {\;^{12}}{C_1}{ \times ^{11}}{C_1}{ \times ^{10}}{C_1} \\ = \frac{{12!}}{{1!\left( {12 - 1} \right)!}} \times \frac{{11!}}{{1!\left( {11 - 1} \right)!}} \times \frac{{10!}}{{1!\left( {10 - 1} \right)!}} \\ = 12 \times 11 \times 10 \\ = 1320 \\\end{aligned} \end{equation} $$

(C) Here again one out of $12$ shirts is chosen.

Since pant is of the same color, only one way of selecting a pant.

Now selection of stockings is one out of $11$ remaining colors.

Total number of selections possible, $$\begin{equation} \begin{aligned} = {\;^{12}}{C_1} \times 1{ \times ^{11}}{C_1} \\ = 132 \\\end{aligned} \end{equation} $$

Question 16. There are $5$ types of beverages in a vending machine.There are $9$ of them, wanting to drink. And each can have only one drink. In how many ways can they chose the beverages?

Solution: $$n = 5\quad {\text{and}}\quad r = 9$$

Thus total number of ways is, $$\begin{equation} \begin{aligned} = {\;^{5 + 9 - 1}}{C_9} \\ = {\;^{13}}{C_9} \\ = \frac{{13!}}{{9!\left( {13 - 9} \right)!}} \\ = 715 \\\end{aligned} \end{equation} $$

Question 17. Consider an $n$-sided polygon, where $n>3$

(A) How many diagonals are there in an $n$-sided polygon?

(B) How many triangles can be formed using the vertices of the polygon?

(C) How many of these have one or two sides in common with the polygon?

Solution:

(A) Diagonals are formed using two points.

Thus $^{n}{C_2}$ ways are there.

Out of these $n$ are sides of the polygon, as selection of two points include adjacent points also.

Thus total number of diagonals that can be formed in an $n$ -sided polygon is, $$ = {\;^n}{C_2} - n$$

(B) There are three points required for a triangle.

Thus the possible number of triangles is, $${ = ^n}{C_3}$$

(C) One side common with the polygon.

Then two points are included always and the adjacent points are excluded.

Thus there are $(n-4)$ points.

The number of triangles formed with one side common is, $$ = n \times {\;^{n - 4}}{C_1}$$

Two side common. The three consecutive points are taken.

Thus total number of such triangles $ = n$.

Question 18. A student is to answer exactly $11$ questions from $15$ , in an examination. Given that she must answer atleast three out of the first six questions, in how many ways can she answer?

Solution: Let the set of questions be split as $6 + 9$.

Atleast $3$ must be answered out of the $6$ and rest from the $9$ questions.

When she attends $3$ from the first six, then the possible selections is, $$ = {\;^6}{C_3}{ \times ^9}{C_8}$$

$4$ from the first six, $$ = {\;^6}{C_4}{ \times ^9}{C_7}$$

$5$ from the first six, $$ = {\;^6}{C_5}{ \times ^9}{C_6}$$

$6$ from the first six, $$ = {\;^6}{C_6}{ \times ^9}{C_5}$$

Thus total such selections is sum of all possible selections, $$ = \left( {^6{C_3}{ \times ^9}{C_8}} \right) + \left( {^6{C_4}{ \times ^9}{C_7}} \right) + \left( {^6{C_5}{ \times ^9}{C_6}} \right) + \left( {^6{C_6}{ \times ^9}{C_5}} \right)$$