Q. 334.0( 9 Votes )

How many four digit natural numbers are not exceeding 4321 can be formed with the digits 1, 2, 3 and 3 if the digits can repeat?

Answer :

We have to find the possible number of four digit numbers which are less than 4321 and are formed with the numbers 1, 2, 3, 4 when repetition of digits is allowed.

We will use the concept of multiplication because there are four sub jobs dependent on each other because the number 4321 cannot be exceeded and we have to look at repetitions carefully.


First, let us consider all the cases in which we do not assign 4 on the thousand’s place which means we will only assign 1, 2, 3 only and repetition in all other places.


The number of ways in which we can assign values to four digit number, but the thousand’s place will not be assigned 4 when repetition is allowed with the help of given data is 3 × 4 × 4 × 4 = 192.


Now we will consider the case when at thousand’s place there is only 4 and nothing else.


So there will be only one choice on thousand’s place that is 4 , there will be two choices on hundred’s place because there will not be 3 and 4 because then our number will become greater than 4321 and 3 will be dealt later by fixing it in hundred’s place like 4 is fixed in this case. The rest places have all the choices.


The number of ways in which we can assign values to the four-digit number and the thousand’s place is only assigned 4 but hundred’s place is not assigned 3 when repetition is allowed with the help of given data is 1 × 2 × 4 × 4 = 32.


Now we will consider the case when thousand’s place is fixed by 4 and hundred’s place is fixed by 3, ten’s place is fixed by 1.


There is one choice each on thousand’s and hundred’s place, but on ten’s place also there is one choice which is 1 , so a number of choices on one's place will be four.


The number of ways in which we can assign values to four-digit number and the thousand’s place is only assigned 4, hundred’s place is only assigned 3, ten’s place is only assigned 1 when repetition is allowed with the help of given data is 1 × 1 × 1 × 4 = 4.


Now we will consider the case when thousand’s place is fixed by 4 and hundred’s place is fixed by 3, ten’s place is fixed by 2.


There is one choice each on thousand’s and hundred’s place, but on ten’s place also there is one choice which is 2 , so a number of choices on one's place will be one.


So the number coming to my mind will be 4321 , which is one number.


Hence total numbers occurring are 192+32+4+1 = 229.


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