Word Problems

Question 11

Question:   In how many ways can a picnic party of 16 people be conveyed in a van, if the first holds not more than 8 and the second one holds not more than 10 people?

Answer:    16 people can be conveyed by

i. Taking 8 peopel in first van and 8 people in second van.
This can be done in ¹⁶C₈ x 1 = ¹⁶C₈ ways.

ii. Accommodating 7 people in first and then 9 people in second van.
This can be done in ¹⁶C₇ x 1 = ¹⁶C₇ ways.

iii. Taking 6 people in first and 10 people in second van.
This can be done in ¹⁶C₆ x 1 = ¹⁶C₆ ways.

\The total number of ways = ¹⁶C₈ + ¹⁶C₇ +¹⁶C₁₀
= 12870 + 11440 + 8008
= 32318

Question 12

Question:   Find the number of lines that can be drawn through 12 points of which 5 are collinear.

Answer:    To draw a line, we need two distinct points.
The number of lines that can be drawn through 12 points is
There are 5 points which are collinear i.e., we can draw only one line through them.
Number of lines that could have been drawn through 5 points
= ⁵C₂ = 10
\ Total number of lines = 66 - 10 - 1 = 57

Question 13

Question:   Find the number of triangles that can be drawn through 12 distinct points of which 5 are collinear.

Answer:    To draw a triangle, we require three non-collinear points.
Hence, the number of triangles that can be drawn through 12 points
= ¹²C₃ = 220
But there are 5 points which are collinear.
Hence, we cannot draw any triangle through 5 points.
We can draw ⁵C₃ triangles through them.
\Total number of triangles = ¹²C₃ - ⁵C₃ = 220 - 10 = 210

Question 14

Question:   From 4 officers and 8 jawans, how many ways can 6 be chosen
i. To include exactly one officer ?
ii. To include at least one officer ?

Answer:   
The number of ways of choosing 5 jawans from 8
\Total number of ways of choosing an officer and 5 jawans

ii.

Question 15

Question:   In a plane, there are n lines, no two of which are parallel and no three are concurrent. How many triangles can be formed with their points of intersection on their vertices?

Answer:    The number of points to take as vertices number of intersection of n

But on each line there are (n-1) collinear points.
(which are the points of intersection of the line with (n-1) lines)

The number of selections of three points for each set of (n-1) collinear points, no triangle can be drawn i.e., ^n-1C₃ triangles are not obtained.
\The required number of triangles

Question 16

Question:   Six X's have to be placed in the squares of the figure shown, such that each row contains at least one X. In how many different ways can this be done?

Answer:    All X's are identical, we have to select 6 squares out of 8. Without
But there are two possibility which are to be excluded
(i) 1st row - 2, 2nd row - 4
(ii) 3rd row -2 , 2nd row - 4

In this case either 1st row is empty or second row is empty. But according to the problem, no row should be empty.

\Required number of selections = 28 - 2 = 26

Question 17

Question:   Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. Determine the number of ways in which the sitting arrangements can be made.

Answer:    9 seats are there on either side of the table. 4 guests desire to sit on one side and 3 others on the opposite side. Thus, we have to choose 5 guests out of 11 who can sit along the four guests already occupying seats on one particular side. The number of ways in which 5 guests out of 11 can be chosen is given by ¹¹C₅. For any one of the above combinations, the number of ways in which the seats can be arranged on the other side will be 9! ways. Thus, for two sides of the table, the number of ways will be (9!)². Hence, for ¹¹C₅ ways of choosing 5 guests out of 11, the total number of arrangements will be (9!)² x ¹¹C_5.

Question 18

Question:   How many three digit numbers can be formed without using the digits 0,2,3,5 and 7?

Answer:    Since, we are not to use the digits 0,2,3,5,and 7, we are left to use the digits 1,4,6,8 and 9, clearly repetition of digits is allowed. Unit's place can be filled in 5 ways. Ten's place can be filled in 5 ways. Hundredth's place can be filled in 5 ways.



By fundamental principle of counting, total three digits numbers are
5 x 5 x 5 = 125.

Question 19

Question:   How many four digit numbers can be formed using the digits 0,2,3,5 and 7, repetition being allowed.

Answer:   
In this problem, we shall begin filling the places from thousand's place.
Since, 0 cannot occupy thousand's place, this place can be filled in 4 ways (First Digit).
Since, 0 can occupy hundred's place, this place can be filled in 5 ways (Second Digit).
Ten's place can be filled in 5 ways (third digit).
Unit's place can also be filled in 5 ways (Fourth digit).

By fundamental principle of counting, the number of four digit numbers are 4 x 5 x 5 x 5 = 500

Question 20

Question:   A team consists of 7 boys and 5 girls and the other team consists of 6 boys and 4 girls. How many single matches can be arranged between the two teams when a boy plays against a boy and a girl against a girl.

Answer:    Since, a boy plays against a boy only, the number of single matches between boys of two teams = 7 x 6 = 42
Similarly, number of single matches between girls of two teams
Total number of events (number of single matches) = 42 + 20 = 62
( The two events does not depend on each other)