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{\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,}} \\

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\[

{\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{.}} \\

\\

\]

{\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{.}} \\

\\

\]

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