How-many-six-digit-numbers-contain-exactly-four-different-digits-

Question Number 21693 by Tinkutara last updated on 01/Oct/17

H o w m a n y s i x - d i g i t n u m b e r s c o n t a i n

e x a c t l y f o u r d i f f e r e n t d i g i t s ?

Commented by mrW1 last updated on 02/Oct/17

9 numbers with exactly 1 distinct digits

2511 numbers with exactly 2 distinct digits

58320 numbers with exactly 3 distinct digits

294840 numbers with exactly 4 distinct digits

408240 numbers with exactly 5 distinct digits

136080 numbers with exactly 6 distinct digits

Σ 900000 numbers

Commented by Tinkutara last updated on 02/Oct/17

Only 4^{th} line is needed . Can you explain

how you get it ?

Commented by mrW1 last updated on 05/Oct/17

for k = 1 \dots 6

\frac{9}{10} \times C_{k}^{10} \times k! \times {_{k}^{6}}

k = 1 : \Rightarrow \frac{9}{10} \times 10 \times 1 \times 1 = 9

k = 2 : \Rightarrow \frac{9}{10} \times 45 \times 2 \times 31 = 2511

k = 3 : \Rightarrow \frac{9}{10} \times 120 \times 6 \times 90 = 58320

k = 4 : \Rightarrow \frac{9}{10} \times 210 \times 24 \times 65 = 294840

k = 5 : \Rightarrow \frac{9}{10} \times 252 \times 120 \times 15 = 408240

k = 6 : \Rightarrow \frac{9}{10} \times 210 \times 720 \times 1 = 136080

Commented by mrW1 last updated on 05/Oct/17

explanation :

to select k digits from 10 (incl . 0) there

are C_{k}^{10} ways

to divide 6 positions into k groups there

are {_{k}^{6}} ways, where {} means stirling

number of second kind

to arrange k digits there are k! ways

since digit 0 is not allowed in the first

position, the result should be reduced

to \frac{9}{10}

\Rightarrow \frac{9}{10} \times C_{k}^{10} \times k! \times {_{k}^{6}}

I worked out this solution by myself,

so I^{'} m not sure if {it}^{'} s correct . pls check .

Commented by Tinkutara last updated on 05/Oct/17

I {don}^{'} t know any sterling function .

Commented by mrW1 last updated on 05/Oct/17

the number of ways to divide n

distinct objects into k groups with at

least one object in each group is :

{_{k}^{n}} = \frac{1}{k!} \underset{j = 0}{\sum^{k}} {(- 1)}^{k - j} (_{j}^{k}) j^{n}

{} is called stirling number of 2 . kind

Commented by mrW1 last updated on 05/Oct/17

Commented by Tinkutara last updated on 05/Oct/17

Thank you very much Sir!

Commented by mrW1 last updated on 06/Oct/17