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Question Number 198242 by Tawa11 last updated on 15/Oct/23
How many numbers with a maximum of 5 digits, greater than 4000, can be formed with the digits 2, 3, 4, 5, 6; if repetition is allowed for 2 and 3 only?
How many numbers with a maximum of 5 digits,
greater than 4000, can be formed with the digits
2, 3, 4, 5, 6; if repetition is allowed for 2 and 3 only?
Commented by necx122 last updated on 15/Oct/23
In those days there was a way Mr. W tackled questions like this. Unfortunately, I dont remember but I wish he attempts also with his method(s).
$${In}\:{those}\:{days}\:{there}\:{was}\:{a}\:{way}\:{Mr}.\:{W} \\ $$$${tackled}\:{questions}\:{like}\:{this}.\:{Unfortunately}, \\ $$$${I}\:{dont}\:{remember}\:{but}\:{I}\:{wish}\:{he}\:{attempts} \\ $$$${also}\:{with}\:{his}\:{method}\left({s}\right). \\ $$
Commented by mr W last updated on 22/Oct/23
thanks that you noticed and like my method! i′ll give a try!
$${thanks}\:{that}\:{you}\:{noticed}\:{and}\:{like}\:{my} \\ $$$${method}!\:{i}'{ll}\:{give}\:{a}\:{try}! \\ $$
Commented by mr W last updated on 22/Oct/23
an alternative solution see Q198576
$${an}\:{alternative}\:{solution}\:{see}\:{Q}\mathrm{198576} \\ $$
Answered by MM42 last updated on 22/Oct/23
4 digits {a,b,c,d} : 3×4×3×2=72 {a,a,b,c} : ((2),(1) )× ((4),(2) )×((4!)/(2!))=144 {a,a,b,b} : ((2),(2) )×((4!)/(2!×2!))=6 {a,a,a,b} : ((2),(1) )× ((4),(1) )×((4!)/(3!))=32 {a,a,a,a} : 2 5 digits {a,b,c,d,e} : 5!=120 {a,a,b,c,d} : ((2),(1) ) × ((4),(3) )×((5!)/(2!))=480 {a,a,b,b,c} : ((2),(2) ) × ((3),(1) )×((5!)/(2!×2!))=180 {a,a,a,b,c} : ((2),(1) ) × ((4),(2) )×((5!)/(3!))=240 {a,a,a,b,b} : ((2),(2) ) ×((5!)/(2!×3!))=10 {a,a,a,a,b} : ((2),(1) ) × ((4),(1) )×((5!)/(4!))=40 {a,a,a,a,a} : 2 ⇒ans=1328 ✓
$$ \\ $$$$\mathrm{4}\:{digits} \\ $$$$\left\{{a},{b},{c},{d}\right\}\::\:\:\mathrm{3}×\mathrm{4}×\mathrm{3}×\mathrm{2}=\mathrm{72} \\ $$$$\left\{{a},{a},{b},{c}\right\}\::\:\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix}×\frac{\mathrm{4}!}{\mathrm{2}!}=\mathrm{144} \\ $$$$\left\{{a},{a},{b},{b}\right\}\::\:\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}×\frac{\mathrm{4}!}{\mathrm{2}!×\mathrm{2}!}=\mathrm{6} \\ $$$$\left\{{a},{a},{a},{b}\right\}\::\:\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{1}}\end{pmatrix}×\frac{\mathrm{4}!}{\mathrm{3}!}=\mathrm{32} \\ $$$$\left\{{a},{a},{a},{a}\right\}\:\::\:\mathrm{2} \\ $$$$\mathrm{5}\:{digits} \\ $$$$\left\{{a},{b},{c},{d},{e}\right\}\::\:\:\mathrm{5}!=\mathrm{120} \\ $$$$\left\{{a},{a},{b},{c},{d}\right\}\::\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{3}}\end{pmatrix}×\frac{\mathrm{5}!}{\mathrm{2}!}=\mathrm{480} \\ $$$$\left\{{a},{a},{b},{b},{c}\right\}\::\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\:×\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix}×\frac{\mathrm{5}!}{\mathrm{2}!×\mathrm{2}!}=\mathrm{180} \\ $$$$\left\{{a},{a},{a},{b},{c}\right\}\::\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix}×\frac{\mathrm{5}!}{\mathrm{3}!}=\mathrm{240} \\ $$$$\left\{{a},{a},{a},{b},{b}\right\}\::\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\:×\frac{\mathrm{5}!}{\mathrm{2}!×\mathrm{3}!}=\mathrm{10} \\ $$$$\left\{{a},{a},{a},{a},{b}\right\}\::\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{1}}\end{pmatrix}×\frac{\mathrm{5}!}{\mathrm{4}!}=\mathrm{40} \\ $$$$\left\{{a},{a},{a},{a},{a}\right\}\::\:\mathrm{2} \\ $$$$ \\ $$$$\Rightarrow{ans}=\mathrm{1328}\:\checkmark \\ $$$$\:\: \\ $$
Commented by Tawa11 last updated on 15/Oct/23
God bless you sir. I appreciate your time.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}. \\ $$
Commented by mr W last updated on 22/Oct/23
miss Tawa: is this result correct?
$${miss}\:{Tawa}: \\ $$$${is}\:{this}\:{result}\:{correct}? \\ $$
Commented by mr W last updated on 22/Oct/23
MM42 sir: please recheck your solution! 4−digit numbers should be larger than 4000. this is not considered in all of your combinations. i think the correct answer should be: 4 digits {a,b,c,d} : 3×4×3×2=72 {b,a,a,c} : ((2),(1) )× ((4),(2) )×3×3×3=54 {a,a,b,b} : ((2),(2) )×((4!)/(2!×2!))=6 {b,a,a,a} : ((2),(1) )× ((3),(1) )×((4!)/(3!))=6 {a,a,a,a} : 2 4 digit numbers: 72+54+6=132 5 digits {a,b,c,d,e} : 5!=120 {a,a,b,c,d} : ((2),(1) ) × ((4),(3) )×((5!)/(2!))=480 {a,a,b,b,c} : ((2),(2) ) × ((3),(1) )×((5!)/(2!×2!))=90 {a,a,a,b,c} : ((2),(1) ) × ((4),(2) )×((5!)/(3!))=240 {a,a,a,b,b} : ((2),(2) ) ×((5!)/(2!×3!))×2=20 {a,a,a,a,b} : ((2),(1) ) × ((4),(1) )×((5!)/(4!))=40 {a,a,a,a,a} : 2 5 digit numbers: 120+480+90+240+20+40+2 =992 totally: 132+992=1124 ✓
$${MM}\mathrm{42}\:{sir}: \\ $$$${please}\:{recheck}\:{your}\:{solution}! \\ $$$$\mathrm{4}−{digit}\:{numbers}\:{should}\:{be}\:{larger} \\ $$$${than}\:\mathrm{4000}.\:{this}\:{is}\:{not}\:{considered}\:{in} \\ $$$${all}\:{of}\:{your}\:{combinations}. \\ $$$$ \\ $$$${i}\:{think}\:{the}\:{correct}\:{answer}\:{should}\:{be}: \\ $$$$ \\ $$$$\mathrm{4}\:{digits} \\ $$$$\left\{{a},{b},{c},{d}\right\}\::\:\:\mathrm{3}×\mathrm{4}×\mathrm{3}×\mathrm{2}=\mathrm{72} \\ $$$$\left\{{b},{a},{a},{c}\right\}\::\:\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}×\cancel{\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix}}×\mathrm{3}×\mathrm{3}×\mathrm{3}=\mathrm{54} \\ $$$$\left\{\cancel{{a},{a},{b},{b}}\right\}\::\:\cancel{\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}×\frac{\mathrm{4}!}{\mathrm{2}!×\mathrm{2}!}=\mathrm{6}} \\ $$$$\left\{{b},{a},{a},{a}\right\}\::\:\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}×\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix}×\cancel{\frac{\mathrm{4}!}{\mathrm{3}!}}=\mathrm{6} \\ $$$$\cancel{\left\{{a},{a},{a},{a}\right\}\:\::\:\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{4}\:{digit}\:{numbers}:\:\mathrm{72}+\mathrm{54}+\mathrm{6}=\mathrm{132} \\ $$$$ \\ $$$$\:\: \\ $$$$\mathrm{5}\:{digits} \\ $$$$\left\{{a},{b},{c},{d},{e}\right\}\::\:\:\mathrm{5}!=\mathrm{120} \\ $$$$\left\{{a},{a},{b},{c},{d}\right\}\::\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{3}}\end{pmatrix}×\frac{\mathrm{5}!}{\mathrm{2}!}=\mathrm{480} \\ $$$$\left\{{a},{a},{b},{b},{c}\right\}\::\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\:×\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix}×\frac{\mathrm{5}!}{\mathrm{2}!×\mathrm{2}!}=\mathrm{90} \\ $$$$\left\{{a},{a},{a},{b},{c}\right\}\::\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix}×\frac{\mathrm{5}!}{\mathrm{3}!}=\mathrm{240} \\ $$$$\left\{{a},{a},{a},{b},{b}\right\}\::\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\:×\frac{\mathrm{5}!}{\mathrm{2}!×\mathrm{3}!}×\mathrm{2}=\mathrm{20} \\ $$$$\left\{{a},{a},{a},{a},{b}\right\}\::\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{1}}\end{pmatrix}×\frac{\mathrm{5}!}{\mathrm{4}!}=\mathrm{40} \\ $$$$\left\{{a},{a},{a},{a},{a}\right\}\::\:\mathrm{2} \\ $$$$ \\ $$$$\mathrm{5}\:{digit}\:{numbers}:\: \\ $$$$\:\:\:\:\mathrm{120}+\mathrm{480}+\mathrm{90}+\mathrm{240}+\mathrm{20}+\mathrm{40}+\mathrm{2} \\ $$$$\:\:\:\:=\mathrm{992} \\ $$$$ \\ $$$${totally}:\:\mathrm{132}+\mathrm{992}=\mathrm{1124}\:\checkmark \\ $$
Commented by mr W last updated on 22/Oct/23
this result can also be obtained using generating function method, see Q198576
$${this}\:{result}\:{can}\:{also}\:{be}\:{obtained} \\ $$$${using}\:{generating}\:{function}\:{method}, \\ $$$${see}\:{Q}\mathrm{198576} \\ $$

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