Permutations & Combinations – I

Here given digits are 10 in number.
Therefore, number of different numbers of 4 digits which can be formed from the 10 digits = ¹⁰P₄. These contain those numbers also which begin with 0 and the number of such numbers = the numbers of ways in which 4-1 i.e. 3 digits can be chosen out of the remaining 10-1 i.e. 9 digits = ⁹P₃
Therefore, required number of 4-digit numbers = ¹⁰P₄–⁹P₃
= (10*9*8*7) -(9*8*7) = (9*8*7) (10-1) = 4536

 

Alternatively

The thousandth place can be filled in 9 ways (any of the 9 digits from 1-9), the hundredth place can be filled in 9 digits (any of the remaining 8 digits plus 0), the tens place can be filled in 7 ways (any of the remaining 7 digits) and the unit’s place can be filled in 6 ways (any of the remaining 6 digits).
Applying the fundamental principle of counting, required number of 4-digit numbers = 9*9*8*7 = 4536

There are ten digits only viz. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 which can be used to form various numbers. Therefore, number of 9-digit numbers formed out of these 10 digits = ¹⁰P₉.
But these include those numbers also which begin with 0 and so are to be excluded as these are not of nine digits.
The number of such numbers having 0 on the extreme left = the number of ways in which remaining eight places are to be filled from the remaining nine digits = ⁹P₈
Therefore, required number = ¹⁰P₉ – ⁹P₈
= 10!-9!
=10*9!-9! = (10-1)9!
=9*9! = 3265920

 

Alternatively
The nine digits have to occupy the following places, without repetition:
A B C D E F G H I
Starting from the left, A can be filled in 9 ways (any of the digits from 1-9), B can be filled in 9 ways (any 8 of the remaining digits from 1-9 plus 0), C can be filled in 8 ways (any 8 of the remaining digits from 0-9), D can be filled in 7 ways (any 7 of the remaining digits from 0-9) etc.
Applying the fundamental principle of counting, required number of ways = 9*9*8*7*6*5*4*3*2 = 3265920

 

i. Here we are to use all the six given digits viz. 4, 5, 6, 7, 8, 9
Therefore, required number of 6-digit numbers = 6! = 720
ii. Now out of these numbers, those which are divisible by 5 have 5 as the last digit and so we have to fill the remaining five places by the remaining 5 given digits and this can be done in 5! ways.
Hence the required number of those numbers which are not divisible by 5
=6!-5! = 720-120 = 600

 

i. The numbers formed by using all the six digits at a time are
⁶P₆ i.e. 6! i.e. 720
But these contain those numbers also which begin with zero and thus are not of six digits. Those numbers having 0 at the first place on the left are ⁵P₅ i.e. 5! i.e. 120 in number.
Therefore, required number of 6-digit numbers which can be formed with the given digits = 720-120 = 600

 

ii.  Also, out of these numbers, the numbers which are divisible by 10 have 0 in the unit’s place i.e. on the extreme right. Thus, we have to arrange the remaining five digits in first five places of the required numbers and this can be done in ⁵P₅ i.e. 5! i.e. 120 ways. Therefore, required 6-digit numbers which are divisible by 10 are 120.

Here, as the numbers to be formed must be divisible by 5, so the last digit of each number must be 5. Again, these numbers must lie between 3000 and 4000 so every such number must be of 4 digits and begin with 3. Thus, we conclude that each number to be formed must be of 4 digits and begin with 3 and end with 5. Now there are only two places in between 3 and 5 which are to be filled with remaining four given digits, as no digit is to be repeated and this can be done is ⁴P₂ ways.
Hence the required number
=⁴P₂ = 4*3 = 12

Here we are given six digits and we are to form numbers of six digits out of these, excluding those numbers which begin with 0 and this can be done in
⁶P₆ – ⁵P₅ i.e. 6!-5! i.e. 720-120 i.e. 600 ways.

Numbers which are divisible by 5, must have either 5 or 0 as the last digit on the extreme right. Now the number of 6-digit numbers having 0 as the last digit on the extreme right = number of ways in which remaining five places are to be filled from the remaining five digits = 5!
Similarly, the number of 6-digit numbers having 5 as the last digit on the extreme right = 5!
But these contain those numbers also which begin with zero and end with 5 and are of six digits and are to be excluded.
Their number = number of ways in which remaining four places are to be filled from the remaining four digits = 4!
Therefore, the number of 6-digit numbers having 5 as the last digit
= 5! – 4!

Therefore, required number of these 6-digit numbers which are divisible by 5 = 5! + (5! – 4!) = 120+(120 – 24) = 216

Every odd number formed by the given five digits will have either 3 or 5 as the last digit on the extreme right i.e. in the unit place.
Now if 3 is fixed at the unit’s place, then remaining four places on the left can be filled in 4! i.e. 24 ways. Thus, the number of 5-digit numbers having 3 at unit place = 24. But these include those numbers also which begin with zero and their number = ³P₃ or 3!
= 6.
But the numbers beginning with zero will not be of five digits, so these are to be excluded. Hence the number of 5-digit numbers having 3 at unit place = 24-6 = 18
Similarly, the number of 5-digit numbers having 5 at unit place = 24-6 = 18
Therefore, the required total number of odd numbers having five digits from the given digits = 18+18 = 36

We are given five digits 1, 2, 3, 4, 5
i. Even numbers formed by the given digits viz. 1, 2, 3, 4, 5 will have either 2 or 4 in the unit’s place. If we fix 2 in the unit’s place then remaining four digits can be placed in the remaining four places in 4! i.e. 24 ways. Similarly, the number of numbers which can be formed from the given digits with 4 in the unit’s place is 4! i.e. 24.

Therefore, total number of even numbers using all the given digits = 24+24 = 48

ii. Numbers which are less than 40000 will have either 1 or 2 or 3 in the first place on the left. If 1 is fixed at the first place on the right, then remaining four digits can be filled in the remaining four places (as all the digits are to be used) in 4! or 24 ways as in case i above. Similarly, the number of numbers beginning with 2 or 3 will be 24 each.

Therefore, required number in this case = 24+24+24 = 72

 

i. All those numbers which lie between 2000 and 3000 are of 4 digits and will begin with 2.
Hence if 2 is fixed at the first place from left, then there will be three places left to be filled by the remaining four digits and this can be done in ⁴P₃ ways i.e. 4*3*2 i.e. 24 ways. Hence the required number in this case = 24

ii. The numbers which are greater than 70000 will be of five digits and will begin with 7. Therefore, if we fix 7 at the first place on the left, then we would be left with four places to be filled by the remaining four digits and this can be done in ⁴P₄ i.e. 4! i.e. 24 ways.
Therefore, 24 numbers can be formed which are greater than 70000 out of the given digits.

iii In this case we are to find those numbers which are of five digits, less than 30000 and divisible by 2.
Therefore, these numbers must begin with 2 and end with 0.
Hence the first place on the right as well as on the left are fixed and we are left to fill the remaining three places with remaining three digits and this can be done in ³P₃3 i.e. 3! i.e. 6 ways. Hence the required number in this case is 6.