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Question: \( {\text{Show the total number of natural numbers of six digit that can be made with }} \\\ ...

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

Explanation

Solution

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 digits. So, there can be two cases, Case 1: 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   4C1ways=4 ways.  Now we have a set of 6 digits out of which three are same and they can be arranged in   6!3!(same)=120ways.  Hence by fundamental theorem the number of such numbers   = 4 \times120 = 480. Case 2: Out of 6 digits 2 are  same,2 are same,  2 are different     This case occurs only when out of the 4 digits we can select 2 sets of same .  So, for selecting 2 number from 4 dgits that occurs 2 times in 6 digit number is   This can be done in   4C2ways=6 ways.  Now we have a set of 6 digits out of which 2 are same of one kind and 2 of other kind.  They can be arranged in   6!2!(same)2!(same)=7204=180ways.  Hence by fundamental theorem the number of such numbers   = 6 \times180 = 1880 . So total number of numbers that can be made from both case will be 480 + 1080 = 1560. Hence proved NOTE: - Whenever you came 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.   {\text{We have to choose numbers having }}6{\text{ digits and made using all the four digits 1,2,3,4}}{\text{. }} \\\ {\text{This is possible if some of the digits repeat to make }}6{\text{ digits}}. \\\ {\text{So, there can be two cases,}} \\\ {\text{Case }}1:{\text{ Out of 6 digits }}3{\text{ are different}},\;3{\text{ are }}\;{\text{same}}\; \\\ {\text{ This case occurs only when one number out of four will appear three times}}{\text{.}} \\\ {\text{ So, for selecting one number that occurs 3 times in 6 digit number is,}} \\\ {\text{ This can be done in}}\;{\text{ }}{}^4{C_1}ways = 4{\text{ }}ways. \\\ {\text{ Now we have a set of }}6{\text{ digits out of which three are same and they can be arranged in}}\;{\text{ }}\dfrac{{6!}}{{3!(same)}} = 120ways. \\\ {\text{ Hence by fundamental theorem the number of such numbers\; = 4 \times 120 = 480}}{\text{.}} \\\ {\text{Case 2}}:{\text{ Out of 6 digits 2 are\;same}},2{\text{ are same,}}\;2{\text{ are different }}\; \\\ {\text{ This case occurs only when out of the 4 digits we can select 2 sets of same }}{\text{.}} \\\ {\text{ So, for selecting 2 number from 4 dgits that occurs 2 times in 6 digit number is }} \\\ {\text{ This can be done in}}\;{\text{ }}{}^4{C_2}ways = 6{\text{ }}ways. \\\ {\text{ Now we have a set of }}6{\text{ digits out of which 2 are same of one kind and 2 of other kind}}{\text{.}} \\\ {\text{ They can be arranged in}}\;{\text{ }}\dfrac{{6!}}{{2!(same)2!(same)}} = \dfrac{{720}}{4} = 180ways. \\\ {\text{ Hence by fundamental theorem the number of such numbers\; = 6 \times 180 = 1880 }}{\text{.}} \\\ {\text{So total number of numbers that can be made from both case will be 480 + 1080 = 1560}}{\text{.}} \\\ {\text{Hence proved}} \\\ {\text{NOTE: - Whenever you came up with this type of problem then the best way is to break the }} \\\ {\text{problem into different cases and find the value for each case then add them up}}{\text{.}} \\\ \\\