a-6-digit-number-formed-with-the-first-number-is-2-and-contains-4-equal-number-many-number-can-be-made-are-

Question Number 77394 by jagoll last updated on 06/Jan/20

$$\mathrm{a}\:\mathrm{6}\:\mathrm{digit}\:\mathrm{number}\:\mathrm{formed}\: \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{first}\:\mathrm{number}\:\mathrm{is}\:\mathrm{2}\:\mathrm{and}\: \\ $$$$\mathrm{contains}\:\mathrm{4}\:\mathrm{equal}\:\mathrm{number}.\:\mathrm{many} \\ $$$$\mathrm{number}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made}\:\mathrm{are}\: \\ $$

Commented by mr W last updated on 06/Jan/20

the first digit is 2, that′s clear. it must contain 4 equal digits. but what about the remaining two digits? may they the same digit or they must be different digits? example: are 233332 and 233222 valid?

$${the}\:{first}\:{digit}\:{is}\:\mathrm{2},\:{that}'{s}\:{clear}. \\ $$$${it}\:{must}\:{contain}\:\mathrm{4}\:{equal}\:{digits}.\:{but} \\ $$$${what}\:{about}\:{the}\:{remaining}\:{two} \\ $$$${digits}?\:{may}\:{they}\:{the}\:{same}\:{digit}\:{or} \\ $$$${they}\:{must}\:{be}\:{different}\:{digits}? \\ $$$${example}:\:{are}\:\mathrm{233332}\:{and}\:\mathrm{233222} \\ $$$${valid}? \\ $$

Commented by jagoll last updated on 06/Jan/20

$${valid}\:{sir} \\ $$

Commented by jagoll last updated on 06/Jan/20

$${i}\:{got}\:\mathrm{3500}\:{sir}?\:{it}\:{correct}? \\ $$

Commented by mr W last updated on 06/Jan/20

$${i}\:{think}\:{it}'{s}\:{not}\:{correct}\:{sir}. \\ $$

Answered by mr W last updated on 06/Jan/20

case 1: in the remaining 5 digits we have 3 times the digit 2. case 1a: type 2222XX ⇒9×((5!)/(3!2!))=90 case 1b: type 2222XY ⇒C_2 ^9 ×((5!)/(3!))=720 case 2: type 2XXXXY in the remaining 5 digits we have 4 times an other digit than 2. ⇒9×9×((5!)/(4!))=405 totally: 90+720+405=1215 numbers

$${case}\:\mathrm{1}: \\ $$$${in}\:{the}\:{remaining}\:\mathrm{5}\:{digits}\:{we}\:{have} \\ $$$$\mathrm{3}\:{times}\:{the}\:{digit}\:\mathrm{2}.\: \\ $$$${case}\:\mathrm{1}{a}:\:{type}\:\mathrm{2222}{XX} \\ $$$$\Rightarrow\mathrm{9}×\frac{\mathrm{5}!}{\mathrm{3}!\mathrm{2}!}=\mathrm{90} \\ $$$${case}\:\mathrm{1}{b}:\:{type}\:\mathrm{2222}{XY} \\ $$$$\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{9}} ×\frac{\mathrm{5}!}{\mathrm{3}!}=\mathrm{720} \\ $$$${case}\:\mathrm{2}:\:{type}\:\mathrm{2}{XXXXY} \\ $$$${in}\:{the}\:{remaining}\:\mathrm{5}\:{digits}\:{we}\:{have} \\ $$$$\mathrm{4}\:{times}\:{an}\:{other}\:{digit}\:{than}\:\mathrm{2}. \\ $$$$\Rightarrow\mathrm{9}×\mathrm{9}×\frac{\mathrm{5}!}{\mathrm{4}!}=\mathrm{405} \\ $$$$ \\ $$$${totally}: \\ $$$$\mathrm{90}+\mathrm{720}+\mathrm{405}=\mathrm{1215}\:{numbers} \\ $$

Commented by jagoll last updated on 06/Jan/20

$${how}\:{about}\:\mathrm{2}{XXXX}\mathrm{2}? \\ $$

Commented by jagoll last updated on 06/Jan/20

$$=\mathrm{9}×\:\frac{\mathrm{5}!}{\mathrm{4}!}\:=\mathrm{4}\:\mathrm{5} \\ $$

Commented by mr W last updated on 06/Jan/20

this is included in 2XXXXY. X can be one of 9 digits (0,1,3,4,...9) Y can be one of 9 digits (0,1,2,3,4,...9)−X

$${this}\:{is}\:{included}\:{in}\:\mathrm{2}{XXXXY}. \\ $$$${X}\:{can}\:{be}\:{one}\:{of}\:\mathrm{9}\:{digits}\:\left(\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{4},…\mathrm{9}\right) \\ $$$${Y}\:{can}\:{be}\:{one}\:{of}\:\mathrm{9}\:{digits}\:\left(\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},…\mathrm{9}\right)−{X} \\ $$

Commented by jagoll last updated on 06/Jan/20