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-

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?

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).

I n t h o s e d a y s t h e r e w a s a w a y M r . W

t a c k l e d q u e s t i o n s l i k e t h i s . U n f o r t u n a t e l y,

I d o n t r e m e m b e r b u t I w i s h h e a t t e m p t s

a l s o w i t h h i s m e t h o d (s) .

Commented by mr W last updated on 22/Oct/23

thanks that you noticed and like my method! i′ll give a try!

t h a n k s t h a t y o u n o t i c e d a n d l i k e m y

m e t h o d! i^{'} l l g i v e a t r y!

Commented by mr W last updated on 22/Oct/23

a n a l t e r n a t i v e s o l u t i o n s e e Q 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 ✓

4 d i g i t s

{a, b, c, d} : 3 \times 4 \times 3 \times 2 = 72

{a, a, b, c} : (\begin{matrix} 2 \\ 1 \end{matrix}) \times (\begin{matrix} 4 \\ 2 \end{matrix}) \times \frac{4!}{2!} = 144

{a, a, b, b} : (\begin{matrix} 2 \\ 2 \end{matrix}) \times \frac{4!}{2! \times 2!} = 6

{a, a, a, b} : (\begin{matrix} 2 \\ 1 \end{matrix}) \times (\begin{matrix} 4 \\ 1 \end{matrix}) \times \frac{4!}{3!} = 32

{a, a, a, a} : 2

5 d i g i t s

{a, b, c, d, e} : 5! = 120

{a, a, b, c, d} : (\begin{matrix} 2 \\ 1 \end{matrix}) \times (\begin{matrix} 4 \\ 3 \end{matrix}) \times \frac{5!}{2!} = 480

{a, a, b, b, c} : (\begin{matrix} 2 \\ 2 \end{matrix}) \times (\begin{matrix} 3 \\ 1 \end{matrix}) \times \frac{5!}{2! \times 2!} = 180

{a, a, a, b, c} : (\begin{matrix} 2 \\ 1 \end{matrix}) \times (\begin{matrix} 4 \\ 2 \end{matrix}) \times \frac{5!}{3!} = 240

{a, a, a, b, b} : (\begin{matrix} 2 \\ 2 \end{matrix}) \times \frac{5!}{2! \times 3!} = 10

{a, a, a, a, b} : (\begin{matrix} 2 \\ 1 \end{matrix}) \times (\begin{matrix} 4 \\ 1 \end{matrix}) \times \frac{5!}{4!} = 40

{a, a, a, a, a} : 2

\Rightarrow a n s = 1328 ✓

Commented by Tawa11 last updated on 15/Oct/23

God bless you sir. I appreciate your time.

God bless you sir .

I appreciate your time .

Commented by mr W last updated on 22/Oct/23

m i s s T a w a :

i s t h i s r e s u l t c o r r e c t ?

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 ✓

M M 42 s i r :

p l e a s e r e c h e c k y o u r s o l u t i o n!

4 - d i g i t n u m b e r s s h o u l d b e l a r g e r

t h a n 4000 . t h i s i s n o t c o n s i d e r e d i n

a l l o f y o u r c o m b i n a t i o n s .

i t h i n k t h e c o r r e c t a n s w e r s h o u l d b e :

4 d i g i t s

{a, b, c, d} : 3 \times 4 \times 3 \times 2 = 72

{b, a, a, c} : (\begin{matrix} 2 \\ 1 \end{matrix}) \times (\begin{matrix} 4 \\ 2 \end{matrix}) \times 3 \times 3 \times 3 = 54

{a, a, b, b} : (\begin{matrix} 2 \\ 2 \end{matrix}) \times \frac{4!}{2! \times 2!} = 6

{b, a, a, a} : (\begin{matrix} 2 \\ 1 \end{matrix}) \times (\begin{matrix} 3 \\ 1 \end{matrix}) \times \frac{4!}{3!} = 6

{a, a, a, a} : 2

4 d i g i t n u m b e r s : 72 + 54 + 6 = 132

5 d i g i t s

{a, b, c, d, e} : 5! = 120

{a, a, b, c, d} : (\begin{matrix} 2 \\ 1 \end{matrix}) \times (\begin{matrix} 4 \\ 3 \end{matrix}) \times \frac{5!}{2!} = 480

{a, a, b, b, c} : (\begin{matrix} 2 \\ 2 \end{matrix}) \times (\begin{matrix} 3 \\ 1 \end{matrix}) \times \frac{5!}{2! \times 2!} = 90

{a, a, a, b, c} : (\begin{matrix} 2 \\ 1 \end{matrix}) \times (\begin{matrix} 4 \\ 2 \end{matrix}) \times \frac{5!}{3!} = 240

{a, a, a, b, b} : (\begin{matrix} 2 \\ 2 \end{matrix}) \times \frac{5!}{2! \times 3!} \times 2 = 20

{a, a, a, a, b} : (\begin{matrix} 2 \\ 1 \end{matrix}) \times (\begin{matrix} 4 \\ 1 \end{matrix}) \times \frac{5!}{4!} = 40

{a, a, a, a, a} : 2

5 d i g i t n u m b e r s :

120 + 480 + 90 + 240 + 20 + 40 + 2

= 992

t o t a l l y : 132 + 992 = 1124 ✓

Commented by mr W last updated on 22/Oct/23

this result can also be obtained using generating function method, see Q198576

t h i s r e s u l t c a n a l s o b e o b t a i n e d

u s i n g g e n e r a t i n g f u n c t i o n m e t h o d,

s e e Q 198576

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