# Problem 1

How many 3-digits numbers can be formed from the digits 0,1,2,3 and 4 without repetation.(En.md.page no 69)

Interpretation:
3-digit number is to be formed using the numbers 0,1,2,3 and 4
Concepts:

1. when forming a 3-digit number '0' can't occupy the hundred place if it occupies then it becomes 2-digit number
2. how to fill the places wheater to start from the unit place or hundreds place (in case of the given numbers contain '0')
3. if one place is filled with an number then how many digits are left to be filled and how many numbers are there to be left for filling
4. application of fundamental principle of counting

Solution:
The digits in the selection set are {0,1,2,3,4}
A 3-digits number number will have three places i.e unit,tens,hundreds
If the selection set contains '0' then the filling must be started from the highest place, so in this problem we have to fill the hundreds place first.
Hundreds: from the given number of set hundred place can't be filled with '0' because it becomes a 2-digit number. Hence the hundreds place can be filled in by 1,2,3 or by 4 i.e in 4-ways

Tens: After filling a place we left with '0' and other 3 numbers (out of 1,2,3 and 4) so the tens place can be filled in 4 different way

Unit: Now after filling the 2 places the unit place can be filled in 3 different ways

Therefore by counting principle total number of ways that a 3-digit number can be formed using 0,1,2,3 and 4 are 4X4X3=48 numbers

# Problem 2

How many 4-digit numbers can be formed using the digits 1,2,3,7,8 and 9 (repetations not allowed)

1. How many of these are less than 6000?
2. How many of these are even?
3. How many of these end with 7?

Interpretation

1. using the digits 1,2,3,7,8,9 we have to form 4-digit numbers and also find the numbers that are less than 6000
2. using the digits 1,2,3,7,8,9 how many 4-digit even numbers can be formed.
3. using the digits 1,2,3,7,8,9 how many 4-digit numbers can be formed those are ends with 7 or the four digit numbers that are having 7 in unit place.

Concepts:

1. Place value
2. when forming a 4-digit number '0' can't occupy the thousands place if it occupies then it becomes 3-digit number
3. how to fill the places wheather to start from the unit place or thousands place (in case of the given numbers contain '0')
4. if one place is filled with an number then how many digits are left to be filled and how many numbers are there to be left for filling
5. should know the charecteristic of the numbers which are less then 6000
6. one should remember that only the numbers of 4-digits which are less than 6000 are to be find out not the total numbers which are less than 6000 (total numbers less than 6000 can be 4-digits,can be 3-digits,can be 2-digits and can also be of single digit)
7. while forming even numbers one should remember that the number so fomed should end in a even number (0,2,4,6 or 8) or the unit place must be filled with either 0,2,4,6 or by 8 only
8. while forming odd numbers one should remember that the number so fomed should end in a odd number (1,3,5,7 or 9) or the unit place must be filled with either 1,3,5,7 or by 9 only
9. application of fundamental principle of counting

Solution:

• Thousands Place: To form the 4-digit numbers which are less than 6000 the thousands place must be filled with the number which is less than 6 so from the given set of number we can fill it by either 1,2 or by 3 i.e in 3 different ways
• Hundreds Place: after filling one place we left with 5 numbers so hundreds place can be filled with any of these in 5 different ways
• Tens Place: with remaing 4-numbers the tens place can be filled in 4 different ways
• Unit Place: simillarly the unit place can be filled in 3-different ways

Hence by fundamental principle of counting total number of ways =3X5X4X3=180 numbers

• while forming the even number unit place must be filled by either by 2 or 8 from the given set of numbers ( 1,2,3,7,8 and 9)
1. unit place-2 ways
2. thousands place-5 ways
3. Hundreds place -4 ways
4. tens place-3 ways

Hence by fundamental principle of counting total number of ways =2X5X4X3=120 numbers

• numbers ending in 7
1. unit place -1way (i.e unit place can be filled by 7 only)
2. thousands place-5 ways
3. Hundreds place -4 ways
4. tens place-3 ways

Hence by fundamental principle of counting total number of ways =1X5X4X3=60 numbers

# Problem 3

How many 1) lines 2) Triangles can be drawn through 8 points on a circle

Interpretation
Drawing lines and triangle using the 8 points which are on a circumference of a circle (irrespective of the position on circumference , may be equally spaced or not) .students should identify whether the problem is of arrengement or selection of objects.

Concepts

1. Circle
2. Co-linear and non co-linear points
3. Difference bet'n line, ray and segment
4. Points require to draw a triangle
5. Difference bet'n arrangements and selection for eg. If A and B are two objects then AB and BA are different arrangements where as AB and BA are both same when it comes to selection.
6. Expantion of Factorial notation

Solution

• For drawing lines we need atleast two points then selecting 2 points out of 8 given we have

${\displaystyle {^{8}}C_{2}}$ combinations

We have ${\displaystyle {^{n}}C_{r}}$=${\displaystyle {\frac {n!}{(n-r)!r!}}}$

therefore ${\displaystyle {^{8}}C_{2}}$=${\displaystyle {\frac {8!}{(8-2)!2!}}}$

=${\displaystyle {\frac {8!}{6!2!}}}$

=${\displaystyle {\frac {8X7X6X5X4X3X2X1}{6X5X4X3X2X1X2X1}}}$

${\displaystyle {^{8}}C_{2}}$=4X7

${\displaystyle {^{8}}C_{2}}$=28 lines can be drawn

• For drawing a triangle we need 3 non co-linear points hence selecting 3 points out of 8 given we have

$\displaystyle {^{8}}C_{3}$ combinations

We have ${\displaystyle {^{n}}C_{r}}$=${\displaystyle {\frac {n!}{(n-r)!r!}}}$

therefore ${\displaystyle {^{8}}C_{3}}$=${\displaystyle {\frac {8!}{(8-3)!3!}}}$

=${\displaystyle {\frac {8!}{5!3!}}}$

=${\displaystyle {\frac {8X7X6X5X4X3X2X1}{5X4X3X2X1X3X2X1}}}$

${\displaystyle {^{8}}C_{2}}$=8X7

${\displaystyle {^{8}}C_{2}}$=56 triangles can be drawn