# Permutations Combinations L2-3

1918 chairs are arranged around a hexagon such that there are 3 chairs per side. If 18 men are to be seated on these chairs, then how many different arrangements are possible?
Solution
The arrangements vary depends on the first person's position where sits in one of the 3 chairs of one side of the hexagon. He would sit at left, center or right side of the the side of the hexagon. Observe the position of "J". If we fix the first person, remaining persons can sit in the remaining chairs in 17! ways. 18 chairs around hexagon
Hence there $3 \times 17!$ Ways.

20A new flag is to be designed with seven vertical strips using some or all of the colours Blue, Red, Black, White and Yellow. No two adjacent strips should be of the same colour. The number of ways this can be done such that even numbered strips are of the same colour while two consecutive odd numbered strips are of different colours is
Solution
There are 5 ways of choosing a colors for even places.
For ${{1}}^{{{st}}}$ odd place there are 4 ways of choosing a colour from remaining 4 colours for ${{2}}^{{{nd}}}$ odd place there are 3 ways of choosing a colour since two consecutive odd numbered strips should have different colour & similarly there are 3 ways each for ${3^{rd}}$ and ${4^{th}}$ odd place.
Hence, total always $= 5 \times 4 \times 3 \times 3 \times 3 = 540$ ways

21A father purchased dress for his 3 daughters. The dresses are of same color but different size and they are kept in dark room. In how many ways all the 3 daughters will not choose their own dress?
Solution
This is a case of de-arrangements = ${D_n} = 3!\left( {\frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}} - ....} \right)$
So number of ways that none of them chooses their own dress = ${D_3} = 3!\left( {\frac{1}{{2!}} - \frac{1}{{3!}}} \right) = 2$

22A dispatch clerk prepared 10 letters and 10 addressed envelopes. What are the total number of ways that exactly 4 letters went into their intended envelope.
Solution
We can select those 4 letters in ${}^{10}{C_4}$ ways. And the remaining 6 letter go into wrong ones.
So Total ways = ${}^{10}{C_4} \times {D_6}$ = ${}^{10}{C_4} \times 6!\left( {\displaystyle\frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}} - \frac{1}{{5!}} + \frac{1}{{6!}}} \right)$

23A three digit number of the form ‘xyz’ is to be formed such that x > y, x > z, and y < z. How many such numbers are possible if x > 0?
Solution
According to the conditions given, following table can be obtained easily: 24On a ABC, on side AB 5 points are marked, on side BC 4 points are marked, while on side AC 3 points are such that none of the points thus marked are on the vertex, then how many triangles can be drawn by joining these points?
Solution
Out of the given 5, 4 & 3 points on each side, triangles possible are
(i) $^{{5}} {{C}}_{{1}} {{ \times }}^{{4}} {{C}}_{{1}} {{ \times }}^{{3}} {{C}}_{{1}}$ $= 5 \times 4 \times 3 = 60$
(ii) $^{{5}} {{C}}_{{1}} {{ \times }}^{{4}} {{C}}_{{0}} {{ \times }}^{{3}} {{C}}_{{2}}$ $= 5 \times 1 \times 3 = 15$
(iii) $^{{5}} {{C}}_{{1}} {{ \times }}^{{4}} {{C}}_{{2}} {{ \times }}^{{3}} {{C}}_{{0}}$ $= 5 \times 6 \times 1 = 30$
(iv) $^{{5}} {{C}}_{{2}} {{ \times }}^{{4}} {{C}}_{{0}} {{ \times }}^{{3}} {{C}}_{{1}}$ $= 10 \times 1 \times 3 = 30$
(v) $^{{5}} {{C}}_{{2}} {{ \times }}^{{4}} {{C}}_{{1}} {{ \times }}^{{3}} {{C}}_{{0}}$ $= 10 \times 4 \times 1 = 40$
(vi) $^{{5}} {{C}}_{{0}} {{ \times }}^{{4}} {{C}}_{{2}} {{ \times }}^{{3}} {{C}}_{{1}}$ $= 1 \times 6 \times 3 = 18$
(vii) $^{{5}} {{C}}_{{0}} {{ \times }}^{{4}} {{C}}_{{1}} {{ \times }}^{{3}} {{C}}_{{2}}$ $= 1 \times 4 \times 3 = 12$
$= 60 + 15 + 30 + 30 + 40 + 18 + 12 = 205$

25 In a certain there are four states. Each state has four major cities. All the major cities in a state are connected with each other via three different modes of transport namely rail, road and air. But a city is connected a another city via only one mode of transport if it belongs to the other states. What is the total number of different routes constructed among the major cities in the given country?
Solution
All the routes can be classified in two categories: First - routes connecting two cities of same state and Second - routes connecting two cities of different states.
First Category: To calculate this first we can select 1 state out of 4 states in four ways then for that particular state we can select 2 cities out of four cities in 6 ways $\left( {^{{4}} {{C}}_{{2}} } \right)$. Now we know that between these two cities 3 routes are possible. So the total number of routes of this category would be 4 x 6 x 3 = 72.
Second Category: To calculate this first we can select 2 states in 6 ways $\left( {^{{4}} {{C}}_{{2}} } \right)$. Then every city of either state is connected with any city of the other state with one route only and there are 4 cities in either state. So the total number of routes of this category would be 6 x 4 x 4 = 96. Total number of routes = 72 + 96 = 168.

26Five intersecting straight lines are drawn in a plane. What is the minimum and maximum number of points of intersection? What is the maximum number of triangles that can be formed?
Solution
As the lines are intersecting, so the minimum number of intersection points is 1. And maximum number of intersection points can be obtained only if we consider that every pair of two lines gives a distinct intersection point. Thus the maximum number of intersection points would be $^{{5}} {{C}}_{{2}}$ or 10. Again, if we are to calculate the maximum number of triangles can be formed by five intersecting lines we have to consider that every group of three lines produces a distinct triangle and in that way maximum number of triangles would be $^{{5}} {{C}}_{{3}}$ or 10

27 10 points are drawn in a plane such that 3 points are collinear. Then
(i) How many straight lines can be drawn by joining these points?
(ii) How many triangles can be drawn by taking these points as the verticles?
Solution
(i) $^{{{10}}} {{C}}_{{2}} {{ - }}^{{3}} {{C}}_{{2}}$ = 45 - 3 + 1 = 43 lines
(ii) $^{{{10}}} {{C}}_{{3}} {{ - }}^{{3}} {{C}}_{{3}}$ = 120 - 1 = 119 triangles

28Find the number of integral solutions of the equation x + y + z = 20 where x, y, z > 0.
Solution
If x = 1, then possible values of y & z are
1 & 18
2 & 17
..
..

18 & 1
So there are 18 possible values
Similarly for x = 2, there are 17 possible values
Till x = 18 you which there is duly possible values so total ways are
=> 18 + 17 + ............. 1
= $\displaystyle\frac{{18\left( {18 + 1} \right)}}{2}$ = 9 x 19 = 171