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71. Out of eight crew members three particular members can sit only on the left side. Another two particular members can sit only on the right side. Find the number of ways in which the crew can be arranged so that four men can sit on each side.
= 3C1 × 4! × 4!
= 1728
Solution:
Required number of ways,= 3C1 × 4! × 4!
= 1728
72. A man positioned at the origin of the coordinate system. the man can take steps of unit measure in the direction North, East, West or South. Find the number of ways of he can reach the point (5,6), covering the shortest possible distance.
The number of ways in which these steps can be taken is given by:
=
= 462
Solution:
In order to reach (5,6) covering the shortest distance at the same time the man has to make 5 horizontal and 6 vertical steps.The number of ways in which these steps can be taken is given by:
=
= 462
73. There are 6 equally spaced points A, B, C, D, E and F marked on a circle with radius R. How many convex pentagons of distinctly different areas can be drawn using these points as vertices?
Solution:
Since, all the points are equally spaced; hence the area of all the convex pentagons will be same.
74. There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?
If arranging these n objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number.
i.e., number of arrangements =
Let there be exactly one person between the two brothers as stated in the question.
If we consider the two brothers and the person in between the brothers as a block, then there will 17 others and this block of three people to be arranged around a circle.
The number of ways of arranging 18 objects around a circle is in 17! ways.
Now the brothers can be arranged on either side of the person who is in between the brothers in 2! ways.
The person who sits in between the two brothers could be any of the 18 in the group and can be selected in 18 ways.
Therefore, the total number of ways
= 18 × 17! × 2
= 2 × 18!
Solution:
n objects can be arranged around a circle in (n - 1)!If arranging these n objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number.
i.e., number of arrangements =
Let there be exactly one person between the two brothers as stated in the question.
If we consider the two brothers and the person in between the brothers as a block, then there will 17 others and this block of three people to be arranged around a circle.
The number of ways of arranging 18 objects around a circle is in 17! ways.
Now the brothers can be arranged on either side of the person who is in between the brothers in 2! ways.
The person who sits in between the two brothers could be any of the 18 in the group and can be selected in 18 ways.
Therefore, the total number of ways
= 18 × 17! × 2
= 2 × 18!
75. a, b, c, d and e are five natural numbers. Find the number of ordered sets (a, b, c, d, e) possible such that a + b + c + d + e = 64.
i.e., o, o, o, o, o, o . . . . (64 balls).
We have 63 gaps where we can place a wall in each gap, since we need 5 compartments we need to place only 4 walls.
We can do this in 63C4 ways.
Solution:
Let assume that there are 64 identical balls which are to be arranged in 5 different compartments (Since a, b, c, d, e are distinguishable) If the balls are arranged in a rowi.e., o, o, o, o, o, o . . . . (64 balls).
We have 63 gaps where we can place a wall in each gap, since we need 5 compartments we need to place only 4 walls.
We can do this in 63C4 ways.
76. There are five cards lying on the table in one row. Five numbers from among 1 to 100 have to be written on them, one number per card, such that the difference between the numbers on any two adjacent cards is not divisible by 4. The remainder when each of the 5 numbers is divided by 4 is written down on another card (the 6th card) in order. How many sequences can be written down on the 6th card?
The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier).
For each value on the card the remainder can have 3 possible values.
The total number of possible sequences is: 4 × 34
Solution:
The remainder on the first card can be 0, 1, 2 or 3 i.e. 4 possibilities.The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier).
For each value on the card the remainder can have 3 possible values.
The total number of possible sequences is: 4 × 34
77. From a total of six men and four ladies a committee of three is to be formed. If Mrs. X is not willing to join the committee in which Mr. Y is a member, whereas Mr.Y is willing to join the committee only if Mrs Z is included, how many such committee are possible?
(i). Mr. Y is a member
(ii). the ones in which he is not
case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.
Now we are left with (6 - 1) men and (4 - 2) ladies (Mrs. X is not willing to join).
We can choose 1 more in 5+2C1 = 7 ways.
case (ii): If Mr. Y is not a member then we left with (6 + 4 - 1) people.
we can select 3 from 9 in 9C3 = 84 ways.
Thus, total number of ways is 7 + 84 = 91 ways.
Solution:
We first count the number of committee in which(i). Mr. Y is a member
(ii). the ones in which he is not
case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.
Now we are left with (6 - 1) men and (4 - 2) ladies (Mrs. X is not willing to join).
We can choose 1 more in 5+2C1 = 7 ways.
case (ii): If Mr. Y is not a member then we left with (6 + 4 - 1) people.
we can select 3 from 9 in 9C3 = 84 ways.
Thus, total number of ways is 7 + 84 = 91 ways.
78. Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?
If A finishes 1 . . . . . B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways = 5 × 4!
If A finishes 2 . . . . . B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways = 4 × 4!
If A finishes 3rd . . . . . B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways = 3 × 4!
If A finishes 4th . . . . . B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways = 2 × 4!
If A finishes 5th . . . . B has to be 6th and the top 4 positions could be filled in 4! ways.
A cannot finish 6th, since he has to be ahead of B.
Therefore total number of ways:
= 5 × 4! + 4 × 4! + 3 × 4! + 2 × 4! + 4!
= 120 + 96 + 72 + 48 + 24
= 360
Solution:
Two horses A and B, in a race of 6 horses . . . A has to finish before B.If A finishes 1 . . . . . B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways = 5 × 4!
If A finishes 2 . . . . . B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways = 4 × 4!
If A finishes 3rd . . . . . B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways = 3 × 4!
If A finishes 4th . . . . . B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways = 2 × 4!
If A finishes 5th . . . . B has to be 6th and the top 4 positions could be filled in 4! ways.
A cannot finish 6th, since he has to be ahead of B.
Therefore total number of ways:
= 5 × 4! + 4 × 4! + 3 × 4! + 2 × 4! + 4!
= 120 + 96 + 72 + 48 + 24
= 360
79. Jay wants to buy a total of 100 plants using exactly a sum of Rs. 1000. He can buy Rose plants at Rs. 20 per plant or marigold or Sun flower plants at Rs. 5 and Rs. 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?
Let number of marigold plants be b.
Let the number of Sunflower plants be c.
According to question,
20a + 5b + 1c = 1000 - - - - - - (1)
a + b + c = 100 - - - - - - - - - - (2)
Solving the above two equations by eliminating c,19a + 4b = 900
b = = - - - - - - - (3)
b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .
0 < b < 99 - - - - - - - (4)
Substituting (3) in (4),
0 < 225 - < 99
⇒ 225 < - < (99 - 225)
⇒ 4 × 225 > 19a > 126 × 4
⇒ > a > 504
a is the integer between 47 and 27 - - - - - - - - (5)
From (3), it is clear, a should be multiple of 4.
Hence, possible values of a are (28,32,36,40,44)
For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40
Three solutions are possible.
Solution:
Let the number of Rose plants be a.Let number of marigold plants be b.
Let the number of Sunflower plants be c.
According to question,
20a + 5b + 1c = 1000 - - - - - - (1)
a + b + c = 100 - - - - - - - - - - (2)
Solving the above two equations by eliminating c,19a + 4b = 900
b = = - - - - - - - (3)
b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .
0 < b < 99 - - - - - - - (4)
Substituting (3) in (4),
0 < 225 - < 99
⇒ 225 < - < (99 - 225)
⇒ 4 × 225 > 19a > 126 × 4
⇒ > a > 504
a is the integer between 47 and 27 - - - - - - - - (5)
From (3), it is clear, a should be multiple of 4.
Hence, possible values of a are (28,32,36,40,44)
For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40
Three solutions are possible.
80. How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?
12C4 ways.
3 vowels can be selected in 4C3 ways.
Therefore, total number of groups each containing 4 consonants and 3 vowels,
= 12C4 × 4C3Each group contains 7 letters, which can be arranging in 7! ways.
Therefore required number of words,
= 12C4 × 4C3 × 7!
Solution:
4 consonants out of 12 can be selected in,12C4 ways.
3 vowels can be selected in 4C3 ways.
Therefore, total number of groups each containing 4 consonants and 3 vowels,
= 12C4 × 4C3Each group contains 7 letters, which can be arranging in 7! ways.
Therefore required number of words,
= 12C4 × 4C3 × 7!