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 198576 by mr W last updated on 22/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 mr W last updated on 22/Oct/23

Q198242 reposted for alternative solutions

Q 198242 r e p o s t e d f o r a l t e r n a t i v e

s o l u t i o n s

Commented by MM42 last updated on 22/Oct/23

Sir W you are right.i was not careful. thank you for your consideration.

S i r W

y o u a r e r i g h t . i w a s n o t c a r e f u l .

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

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

y o u a r e w e l c o m e s i r!

Commented by nikif99 last updated on 23/Oct/23

A more complicated solution I. four digit numbers determinant (((1st),( 2nd),( 3rd),( 4th),( cases^★ ))) 4or5or6^★ { ((2or3 { ((2or3 { ((2or3 (3×2×2×2=24))),((5or6 (24))) :})),((5or6 { ((2or3 (24))),((6 (12))) :})) :})),((5or6^★ { ((2or3 { ((2or3 (24))),((6^★ (12))) :})),((6 { ((2or3 (12))),(( × (0))) :})) :})) :} ★: the product of all possible cases through a routing of digits. Subtotal 132 II. five digit numbers determinant (((1st),( 2nd),( 3rd),( 4th),( 5th),( cases))) determinant (((2or3 { ((2or3 { ((2or3 { ((2or3 { ((2or3 (32))),((4or5or6 (48))) :})),((4or5or6 { ((2or3 (48))),((5or6 (48))) :})) :})),((4or5or6 { ((2or3 { ((2or3 (48))),((5or6 (48))) :})),((5or6 { ((2or3 (48))),((6 (24))) :})) :})) :})),((4or5or6^★ { ((2or3 { ((2or3 { ((2or3 (48))),((5or6 (48))) :})),((5or6 { ((2or3 (48))),((6 (24))) :})) :})),((5or6^★ { ((2or3 { ((2or3 (48))),((6^★ (24))) :})),((6 { ((2 (12))),((3 (12))) :})) :})) :})) :}))) Subtotal 608 4or5or6^★ { ((2or3 { ((2or3 { ((2or3 { ((2or3 (48))),((5or6 (48))) :})),((5or6 { ((2or3 (48))),((6 (24))) :})) :})),((5or6 { ((2or3 { ((2or3 (48))),((6 (24))) :})),((6 { ((2 (12))),((3 (12))) :})) :})) :})),((5or6^★ { ((2or3 { ((2or3 { ((2or3 (48))),((6 (24))) :})),((6^★ { ((2 (8))),((3 (8))) :})) :})),((6 { ((2 { ((2 (8))),((3 (8))) :})),((3 { ((2 (8))),((3 (8))) :})) :})) :})) :} Subtotal 384 Total 1124 ★: if for exemple 4 is selected initialy, then one of {5,6} is selected next, and finally the last one.

A m o r e c o m p l i c a t e d s o l u t i o n

I . f o u r d i g i t n u m b e r s

\begin{matrix} 1 s t & 2 n d & 3 r d & 4 t h & {c a s e s}^{★} \end{matrix}

4 o r 5 o r 6^{★} {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 (3 \times 2 \times 2 \times 2 = 24) \\ 5 o r 6 (24) \end{cases} \\ 5 o r 6 {\begin{cases} 2 o r 3 (24) \\ 6 (12) \end{cases} \end{cases} \\ 5 o r 6^{★} {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 (24) \\ 6^{★} (12) \end{cases} \\ 6 {\begin{cases} 2 o r 3 (12) \\ \times (0) \end{cases} \end{cases} \end{cases}

★ : t h e p r o d u c t o f a l l p o s s i b l e c a s e s t h r o u g h a r o u t i n g o f d i g i t s .

S u b t o t a l 132

I I . f i v e d i g i t n u m b e r s

\begin{matrix} 1 s t & 2 n d & 3 r d & 4 t h & 5 t h & c a s e s \end{matrix}

\begin{matrix} 2 o r 3 {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 (32) \\ 4 o r 5 o r 6 (48) \end{cases} \\ 4 o r 5 o r 6 {\begin{cases} 2 o r 3 (48) \\ 5 o r 6 (48) \end{cases} \end{cases} \\ 4 o r 5 o r 6 {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 (48) \\ 5 o r 6 (48) \end{cases} \\ 5 o r 6 {\begin{cases} 2 o r 3 (48) \\ 6 (24) \end{cases} \end{cases} \end{cases} \\ 4 o r 5 o r 6^{★} {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 (48) \\ 5 o r 6 (48) \end{cases} \\ 5 o r 6 {\begin{cases} 2 o r 3 (48) \\ 6 (24) \end{cases} \end{cases} \\ 5 o r 6^{★} {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 (48) \\ 6^{★} (24) \end{cases} \\ 6 {\begin{cases} 2 (12) \\ 3 (12) \end{cases} \end{cases} \end{cases} \end{cases} \end{matrix}

S u b t o t a l 608

4 o r 5 o r 6^{★} {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 (48) \\ 5 o r 6 (48) \end{cases} \\ 5 o r 6 {\begin{cases} 2 o r 3 (48) \\ 6 (24) \end{cases} \end{cases} \\ 5 o r 6 {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 (48) \\ 6 (24) \end{cases} \\ 6 {\begin{cases} 2 (12) \\ 3 (12) \end{cases} \end{cases} \end{cases} \\ 5 o r 6^{★} {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 {\begin{cases} 2 o r 3 (48) \\ 6 (24) \end{cases} \\ 6^{★} {\begin{cases} 2 (8) \\ 3 (8) \end{cases} \end{cases} \\ 6 {\begin{cases} 2 {\begin{cases} 2 (8) \\ 3 (8) \end{cases} \\ 3 {\begin{cases} 2 (8) \\ 3 (8) \end{cases} \end{cases} \end{cases} \end{cases}

S u b t o t a l 384

T o t a l 1124

★ : i f f o r e x e m p l e 4 i s s e l e c t e d i n i t i a l y, t h e n o n e o f {5, 6} i s s e l e c t e d n e x t,

a n d f i n a l l y t h e l a s t o n e .

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

method using generating function seems to be camplicated at first glance, but actually this solution method is general and thus very powerful. try to find the number of 7−digit numbers with exactly the same conditions. can you give a solution for this case?

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

s e e m s t o b e c a m p l i c a t e d a t f i r s t

g l a n c e, b u t a c t u a l l y t h i s s o l u t i o n

m e t h o d i s g e n e r a l a n d t h u s v e r y

p o w e r f u l . t r y t o f i n d t h e n u m b e r

o f 7 - d i g i t n u m b e r s w i t h e x a c t l y

t h e s a m e c o n d i t i o n s . c a n y o u g i v e

a s o l u t i o n f o r t h i s c a s e ?

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

Commented by nikif99 last updated on 23/Oct/23

It is quite difficult to manage manually such a tree with 7 levels. Trying another approach, I used self−developped software via Fortran and I found 8.864 cases for 7−digit numbers (and 3.040 for 6−digit). P.S. I do not know what other members see on my solution comment; I personally see only the title in bold (“4−digit numbers”) without analysis with opening curly braces.

I t i s q u i t e d i f f i c u l t t o m a n a g e m a n u a l l y

s u c h a t r e e w i t h 7 l e v e l s .

T r y i n g a n o t h e r a p p r o a c h, I u s e d

s e l f - d e v e l o p p e d s o f t w a r e v i a F o r t r a n

a n d I f o u n d 8.864 c a s e s f o r 7 - d i g i t

n u m b e r s (a n d 3.040 f o r 6 - d i g i t) .

P . S . I d o n o t k n o w w h a t o t h e r m e m b e r s

s e e o n m y s o l u t i o n c o m m e n t;

I p e r s o n a l l y s e e o n l y t h e t i t l e i n b o l d

(“ 4 - d i g i t {n u m b e r s}^{″}) w i t h o u t a n a l y s i s

w i t h o p e n i n g c u r l y b r a c e s .

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

i can also only see the title in bold, nothing follows.

i c a n a l s o o n l y s e e t h e t i t l e i n b o l d,

n o t h i n g f o l l o w s .

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

8864 numbers with 7 digits, this is correct, thanks alot! i also got this result, very easily with my method applying generating function. the result is the coefficient of x^7 term in 7!e^(2x) (1+x)^3 , which is 8864. similarly for 6−digit numbers it is the coefficient of x^6 term in 6!e^(2x) (1+x)^3 , which is 3040.

8864 n u m b e r s w i t h 7 d i g i t s, t h i s i s

c o r r e c t, t h a n k s a l o t! i a l s o g o t t h i s

r e s u l t, v e r y e a s i l y w i t h m y m e t h o d

a p p l y i n g g e n e r a t i n g f u n c t i o n . t h e

r e s u l t i s t h e c o e f f i c i e n t o f x^{7} t e r m

i n 7! e^{2 x} {(1 + x)}^{3}, w h i c h i s 8864 .

s i m i l a r l y f o r 6 - d i g i t n u m b e r s i t i s

t h e c o e f f i c i e n t o f x^{6} t e r m i n

6! e^{2 x} {(1 + x)}^{3}, w h i c h i s 3040 .

Commented by nikif99 last updated on 23/Oct/23

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

n i c e p r e p a r e d!

Answered by mr W last updated on 23/Oct/23

2 and 3 can be used zero time, one time or any times ⇒(1+x+x^2 +x^3 +...) 4, 5 and 6 can only be used zero time or one time ⇒(1+x) 5−digit numbers: number of such numbers is the coefficient of x^5 term in 5!(1+x+(x^2 /(2!))+(x^3 /(3!))+...)^2 (1+x)^3 =5!e^(2x) (1+x)^3 which is 992. that means we can form 992 such 5−digit numbers.

2 a n d 3 c a n b e u s e d z e r o t i m e, o n e

t i m e o r a n y t i m e s \Rightarrow (1 + x + x^{2} + x^{3} + \dots)

4, 5 a n d 6 c a n o n l y b e u s e d z e r o t i m e

o r o n e t i m e \Rightarrow (1 + x)

\underset{―}{5 - d i g i t n u m b e r s :}

n u m b e r o f s u c h n u m b e r s i s t h e

c o e f f i c i e n t o f x^{5} t e r m i n

5! {(1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots)}^{2} {(1 + x)}^{3}

= 5! e^{2 x} {(1 + x)}^{3}

w h i c h i s 992 .

t h a t m e a n s w e c a n f o r m 992 s u c h

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

Commented by Tawa11 last updated on 23/Oct/23

Thanks sir. I appreciate your time, God bless you sir.

Thanks sir .

I appreciate your time, God bless you sir .

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

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

4−digit numbers: the first digit must be 4, 5 or 6. we select at first the first digit, there are 3 ways. then we form 3−digit numbers with the other 4 digits. similarly to above, the number of such numbers is the coefficient of x^3 term in 3!e^(2x) (1+x)^2 , which is 44. so totally we can form 3×44=132 valid 4−digit numbers.

\underset{―}{4 - d i g i t n u m b e r s :}

t h e f i r s t d i g i t m u s t b e 4, 5 o r 6 .

w e s e l e c t a t f i r s t t h e f i r s t d i g i t,

t h e r e a r e 3 w a y s . t h e n w e f o r m

3 - d i g i t n u m b e r s w i t h t h e o t h e r 4

d i g i t s . s i m i l a r l y t o a b o v e, t h e n u m b e r

o f s u c h n u m b e r s i s t h e c o e f f i c i e n t

o f x^{3} t e r m i n 3! e^{2 x} {(1 + x)}^{2}, w h i c h i s

44 . s o t o t a l l y w e c a n f o r m

3 \times 44 = 132 v a l i d 4 - d i g i t n u m b e r s .

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

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

a l l t o g e t h e r : 992 + 132 = 1124 ✓

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