Q1. I have four currency notes valued at Rs 50,100 and 2 notes of 500.In how many ways can I select the
currency note ,so that only one type of note is selected each time?
Solution
We have 3 notes here, 50, 100 and 500.
So I can pick a 50 Rupee note, 100 Rupee note and a 500 Rupee note.
So I have 3 selections here.
I can also pick all three at the same time. (50,100,500)
This would count as one selection.
I can also choose two at a time
Hence, I can take either a 50 and a 100 rupee note, a 100 and a 500 rupee note, and a 50 and a 500 rupee note. Hence, there are 3 more cases here.
Total number of possible selections รข€“3+3+1=7
Q2. In a class there are 27 boys and 15 girls. The teacher wants to select a boys and a girl for the monitor ship of the class. In how many ways can the teacher make this selection?
Solution
The teacher has to perform two jobs
First, he has to select a boy among 27 boys.
Second, he has to select a girl among 15 girls.
The first job can be performed in 27 ways and the second in 15 ways. Therefore, by the fundamental principle of multiplication, the required no. of ways of selection is 27 x 15 = 405.
Q3. A room has 5 doors. In how many ways can a man enter the room through one and come out through a different door?
Solution
The man has to perform two jobs in succession:
1) He has to enter the room, then
2) He has to come out from the room
Clearly, a person can enter the room through anyone of the 5 doors, So he can perform first job in 5 ways. After entering into the room, the man can come out through a different door in 4 ways.
Hence the number of ways in which a man can enter a room through one door and come out through a different door = 5 x 4 = 20
Q4. How many words (with or without meaning) of three distinct letters of the English alphabets are there?
Solution
Here we have to make 3 letter words. That means we have to fill up three places by distinct letters of the English alphabets.
Since there are 26 letters of the English alphabet, the first place can be filled by anyone of these letters. So, there are 26 ways of filling up the first place. Now the second place can be filled up by anyone of the remaining 25 letters. So, there are 25 ways of filling up the second place. After filling up the first two places only 24 letters are left to fill up the third place. So the third place can be filled in 24 ways.
Hence the required number of words = 26 x 25 x 24 = 15600
Q5. In how many ways 5 rings of different types can be worn in 4 fingers?
Solution
The first ring can be worn in any of the 4 fingers.
So there are 4 ways of wearing it.
Similarly each one of the other rings can be worn in 4 ways.
Hence, the required number of ways = 4 x 4 x 4 x 4 x 4 = 45
Q6. How many numbers are there between 100 and 1000 which have exactly one of their digits as 7?
Solution
A number between 100 and 1000 contains 3 digits. So, we have to form 3 digit numbers having exactly one of their digits as7. Such type of numbers can be divided into three types:
1) There numbers that have 7 in the unit place but not in any other place.
2) These numbers that have 7 in the ten's place but not in any other place.
3) These numbers that have 7 in the hundred's place but not in any other ways.
We shall now count these three type of numbers separately.
1) These 3-digit numbers that have 7 in the unit's place but not in any other place.
The hundred's place can have any one of the digits from 0 to 9 except 0 and 7. So hundred's place can be filled in 8 ways. The ten's place can have anyone of the digits from 0 to 9 except 7. So the number of ways the ten's place can be filled is 9. The unit's place has 7. So, it can be filled in only one way.
Thus, there are 8 x 9 x 1 = 72 numbers of the first kind.
2) Those three-digit numbers that have 7 in the ten's place but not in any other place.
The number of ways to fill the hundred's place = 8 [By any one of the digits from
1,2,3,4,5,6,8,9]
The number of ways to fill the ten's place = 1 [By 7 only]
The number of ways to fill the one's place = 9 [By any one of the digits 0,1,2,3,4,5,
6,8,9]
Thus, these are 8 x 1 x 9 = 72 numbers of the second kind.
3) These three digit numbers that have 7 in the hundred's place but not at any other place.
In this case, the hundred's place can be filled only in one way and each of the ten's and one's place can be filled in 9 ways.
So, there are 1 x 9 x 9 = 81 numbers of the third kind.
Hence, the total number of required type of numbers = 72 + 72 + 81 = 225.
Q7. How many numbers are there between 100 and 1000 in which all the digits are distinct?
Solution
We know that a number between 100 and 1000 has three digits. So we have to find all the three digit numbers with distinct digits.
We cannot have 0 at the hundred’s place for 3-digit number. So the hundred’s place can be filled with any of the 9 digits. 1,2,3,…..9. So, there are 9 ways of filling the hundred’s place.
Now, 9 digits are left including 0. So ten’s place can be filled with any of the remaining 9 digits in 9 was. Now the unit’s place can be filled with any of the remaining 8 digits. So, there are 8 ways of filling the unit’s place.
Hence, the total number of required numbers = 9 x 9 x 8= 648.
Q8. How many four digit numbers can be formed using the digits 0,1,2,3,4,5 if
1) repetition of digits is not allowed?
2) repetition of digits is allowed?
Solution
1) In the four digit number 0 cannot appear in the thousand’s place. So, thousand’s place
can be filled in 5 ways (using 1,2,3,4 or 5). Since repetition of digits is not allowed and 0 can be used at hundred’s place, so hundred’s place can be filled in 5 ways. Now, any one of the remaining four digits can be used to fill up ten’s place. So ten’s place can be filled in 4 ways. One’s place can be filled from the remaining three digits in 3 ways.
Hence, the required number of numbers = 5 x 5 x 4 x 3 = 300
2) For a four digit number we have to fill up four places and 0 cannot appear in the thousand’s place can be filled in 5 ways. Since repetition of digits is allowed, so each of remaining three places hundred’s, ten’s and one’s can be filled in 6 ways.
Hence, the required number of numbers = 5 x 6 x 6 x 6 = 1080.
Q9. From a class of 25 students, 10 are to be chosen for an excursion party there are 3 students who decide that either all of them will join or none of them will join. In how many ways can they be chosen?
Now a day's teenager's behavior is highly influenced under peer pressure without considering its pros and cons. Do you think that teachers' suggestion plays an important role in development of students' life?
Solution
We have following cases: (i) Three particular students join: now we have to choose 7 students from remaining 22 students So number of ways = 22C7 (ii) Three particular students do not join. Here we have to choose 10 students from remaining 22 students So number of ways = 22C10 Hence, total number of ways = 
+ 22C10 = 817190 Teachers' suggestion plays a very important role in development of students' life. They mould the students to bring out their skills or improvise them, teach good habits/attitudes and help them to become good citizens of the nation.
+ 22C10 = 817190 Teachers' suggestion plays a very important role in development of students' life. They mould the students to bring out their skills or improvise them, teach good habits/attitudes and help them to become good citizens of the nation.
Q10. How many triangles can be drawn through 10 points in a plane?
Solution
A triangle requires three points to be drawn, hence the total number of triangles is 10 C3
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