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Show the total number of natural numbers of six digits that can be made with digits 1,2,3,4 if all numbers are to appear in the same number at least once in 150.

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Hint: Find out all possible ways of combination i.e with and without repetition of numbers and add all the cases to find the solution.


Complete step-by-step answer:

We have to choose numbers having 6 digits and made using all the four digits 1,2,3,4.

 This is possible if some of the digits repeat to make 6.

 So, there can be two cases,

 Case1: Out of 6 digits 3 are different 3 are same

 This case occurs only when one number out of four will appear three times.

 So, for selecting one number that occurs 3 times in 6 digit number is,

 This can be done in ${\text{ }}{}^4{C_1}$ ways = 4 ways.

 Now we have a set of 6 digits out of which three are same and they can be arranged in $\dfrac{{6!}}{{3!}} = 120ways$.

Hence by fundamental theorem the number of such numbers = 4*120 = 480

 

Case 2: Out of 6 digits 2 sets of 2 numbers are the same and 2 are different. 

This case occurs only when out of the 4 digits we can select 2 sets of the same.

So, for selecting 2 number from 4 digits that occurs 2 times in 6 digit number is 

This can be done in ${\text{ }}{}^4{C_2}$ ways = 6 ways. 

Now we have a set of 6 digits out of which 2 are the same of one kind and 2 of other kinds.

They can be arranged in $\dfrac{{6!}}{{2!*2!}} = \dfrac{{720}}{4} = 180ways. $

Hence by fundamental theorem the number of such numbers = 6*180 = 1880.

So the total number of numbers that can be made from both cases will be $480 + 1080 = 1560$.

Hence proved.


NOTE: - Whenever we come up with this type of problem then the best way is to break the problem into different cases and find the value for each case then add them up.


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