RD Chapter 16 Permutations Ex 16.2 Solutions
Question - 11 : - Given 7 flags of different colours, how many different signals can be generated if a signal requires the use of two flags, one below the other?
Answer - 11 : -
Given: seven flags areavailable and out of which two are needed to make a signal.
From this, we can say,that we have to select two flags out of seven and arrange these two flags toget one signal.
Seven flags ofdifferent colours are available, so first flag can be selected in 7 ways.
Now, the second flagcan be selected from any one of the remaining flags in 6 ways.
Hence, the requirednumber of signals are 7 × 6 = 42.
Question - 12 : - A team consists of 6 boys and 4 girls, and other has 5 boys and 3 girls. How many single matches can be arranged between the two teams when a boy plays against a boy, and a girl plays against a girl?
Answer - 12 : -
Given:
A team consists of 6boys and 4 girls, and other team has 5 boys and 3 girls.
Let team 1 be = 6 boysand 4 girls
Team 2 be = 5 boys and3 girls
Singles matches are tobe played, either a boy plays against a boy, and a girl plays against a girl.
So, number ofways to select a boy from team 1 is 6C1. Similarly,Number of ways to select a boy from team 2 is 5C1
Hence number ofsingles matches between boys is 6C1 × 5C1 =6 × 5 = 30
A number of waysto select a girl from team 1 is 4C1. Similarly,Number of ways to select a girl from team 2 is 3C1
Hence number ofsingles matches between girls is 3C1 × 4C1 =4 × 3 = 12
∴ The total number ofsingle matches are = 30 + 12 = 42
Question - 13 : - Twelve students compete in a race. In how many ways first three prizes be given?
Answer - 13 : -
Given:
Twelve studentscompete in a race.
Number of ways toselect the winner of the first prize is 12C1
Number of ways toselect the winner of the second prize is 11C1 (11,since one student is already given a prize)
Number of ways toselect the winner of the third prize is 10C1 (10,since two students are already given a prize)
Hence, total number ofways is 12C1 × 11C1 × 10C1 =12 × 11 × 10 = 1320.
Question - 14 : - How many A.P.’s with 10 terms are there whose first term is in the set {1, 2, 3} and whose common difference is in the set {1, 2, 3, 4, 5}?
Answer - 14 : -
We know that, each AP consists of a unique first term and a common difference.
So, number of ways to select the first term of a given set is 3C1 = 3
And, number of ways to select a common difference of given set is 5C1 = 5
Hence, total number of AP’s possible are 3C1 × 5C1 = 3 × 5 = 15
Question - 15 : - From among the 36 teachers in a college, one principal, one vice-principal and the teacher-in-charge are to be appointed. In how many ways can this be done?
Answer - 15 : -
Number of ways toappoint one principal, vice-principal and the teacher incharge is equal to thenumber of ways to select the three teachers from 36 members.
So, a total of threepositions are to be appointed.
Number of ways toselect principal is 36C1 = 36
Number of ways toselect vice-principal is 35C1 = 35 (35, sinceone position is already given)
Number of ways toselect teacher in charge is 34C1 (34, since twopositions are already given)
Hence, number of waysto appoint three teachers is 36C1 × 35C1 × 34C1 =36 × 35 × 34 = 42840.
Question - 16 : - How many three-digit numbers are there with no digit repeated?
Answer - 16 : -
Let us assume we havethree boxes.
The first box can befilled with any one of the nine digits (0 not allowed at first place).
So, the possibilitiesare 9C1
The second box can befilled with any one of the nine digits
So the availablepossibilities are 9C1
The third box can befilled with any one of the eight digits
So the availablepossibilities are 8C1
Hence, the totalnumber of possible outcomes are 9C1 × 9C1 × 8C1 =9 × 9 × 8 = 648.
Question - 17 : - How many three-digit numbers are there?
Answer - 17 : -
Let us assume we havethree boxes.
The first box can befilled with any one of the nine digits (zero not allowed at first position)
So, possibilitiesare 9C1
The second box can befilled with any one of the ten digits
So the availablepossibilities are 10C1
Third box can befilled with any one of the ten digits
So the availablepossibilities are 10C1
Hence, the totalnumber of possible outcomes are 9C1 × 10C1 × 10C1 =9 × 10 × 10 = 900.
Question - 18 : - How many three-digit odd numbers are there?
Answer - 18 : -
We know that in oddnumbers, the last digit consists of (1, 3, 5, 7, 9).
Let us assume we havethree boxes.
The first box can befilled with any one of the nine digits (zero not allowed at first position)
So the possibilitiesare 9C1
The second box can befilled with any one of the ten digits
So the availablepossibilities are 10C1
The third box can befilled with any one of the five digits (1,3,5,7,9)
So the availablepossibilities are 5C1
Hence, the totalnumber of possible outcomes are 9C1 × 10C1 × 5C1 =9 × 10 × 5 = 450
Question - 19 : - How many different five-digit number license plates can be made if
(i) the first digit cannot be zero, and the repetition of digits is not allowed,
(ii) the first-digit cannot be zero, but the repetition of digits is allowed?
Answer - 19 : -
(i) We know that zerocannot be the first digit of the license plates. And the repetition of digitsis not allowed.
Let us assume five boxes,now the first box can be filled with one of the nine available digits, so thepossibility is 9C1
Similarly, the secondbox can be filled with one of the nine available digits, so the possibilityis 9C1
The third box canbe filled with one of the eight available digits, so the possibility is 8C1
The fourth boxcan be filled with one of the seven available digits, so the possibilityis 7C1
The fifth box canbe filled with one of the six available digits, so the possibility is 6C1
Hence, the totalnumber of possible outcomes is 9C1 × 9C1 × 8C1 × 7C1 × 6C1 =9 × 9 × 8 × 7 × 6 = 27,216
(ii) We know thatzero cannot be the first digit of the license plates. And the repetition ofdigits is allowed.
Let us assume fiveboxes, now the first box can be filled with one of the nine available digits,so the possibility is 9C1
Similarly, the secondbox can be filled with one of the ten available digits, so the possibilityis 10C1
The third box canbe filled with one of the ten available digits, so the possibility is 10C1
The fourth boxcan be filled with one of the ten available digits, so the possibility is 10C1
The fifth box canbe filled with one of the ten available digits, so the possibility is 10C1
Hence, the totalnumber of possible outcomes is 9C1 × 10C1 × 10C1 × 10C1 × 10C1 =9 × 10 × 10 × 10 × 10 = 90,000
Question - 20 : - How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 7000, if repetition of digits is not allowed?
Answer - 20 : -
The required numbersare greater than 7000.
So, the thousand’splace can be filled with any of the 3 digits: 7, 8, 9.
Let us assume fourboxes, now in the first box can either be one of the three numbers 7, 8 or 9,so there are three possibilities which are 3C1
In the second box, thenumbers can be any of the four digits left, so the possibility is 4C1
In the third box, thenumbers can be any of the three digits left, so the possibility is 3C1
In the fourth box, thenumbers can be any of the two digits left, so the possibility is 2C1
Hence, the totalnumber of possible outcomes is 3C1 × 4C1 × 3C1 × 2C1 =3 × 4 × 3 × 2 = 72.