Question

How many even natural numbers divisible by 5 can be formed with the digits 0, 1, 2, 3, 4, 5, 6 if repetition of digits is not allowed?

Answer 1

$If\ the\ even\ natural\ numbers\ can\ be\ divided\ by\ 5,\ then\ the\ numbers \\ are\ multiples\ of\ 10.\ \therefore\ Their\ unit\ digit\ must\ always\ be\ 0.$

$\\ \\ As\ repetition\ is\ not\ allowed,\ none\ other\ than\ unit\ place\ can \\ have\ the\ value\ of\ 0. \\ \\ One-digit\ even\ natural\ numbers\ which\ are\ multiples\ of\ 5 \\ can't\ be\ formed\ with\ 0,\ 1,\ 2,\ 3,\ 4,\ 5\ and\ 6.\ 0\ can\ be\ even\ number \\ and\ multiple\ of\ 5\ but\ is\ not\ natural\ number. \\ \\ On\ 2-digit\ numbers,\ \bold{6}\ numbers\ (1,\ 2,\ 3,\ 4,\ 5,\ 6)\ can\ be\ tens\ digit \\ and\ only\ \bold{1}\ number\ (0)\ can\ be\ unit\ digit.$

$\\ \\ So\ 6 \times 1 = \bold{6}\ \ 2-digit\ numbers\ can\ be\ formed. \\ \\ On\ 3-digit\ numbers, \\ \\ \bold{6}\ numbers\ can\ have\ hundreds\ digit. \\ \bold{5}\ numbers\ can\ have\ tens\ digit. \\ \bold{1}\ number\ can\ have\ unit\ digit. \\ \\ \therefore\ 6 \times 5 \times 1 = \bold{30}\ \ 3-digit\ numbers.$

$\\ \\ On\ 4-digit\ numbers, \\ \\ \bold{6}\ numbers\ can\ have\ thousands\ digit. \\ \bold{5}\ numbers\ can\ have\ hundreds\ digit. \\ \bold{4}\ numbers\ can\ have\ tens\ digit. \\ \bold{1}\ number\ can\ have\ unit\ digit. \\ \\ \therefore\ 6 \times 5 \times 4 \times 1 = \bold{120}\ \ 4-digit\ numbers.$

$\\ \\ On\ 5-digit\ numbers, \\ \\ \bold{6}\ numbers\ can\ have\ ten\ thousands\ digit. \\ \bold{5}\ numbers\ can\ have\ thousands\ digit. \\ \bold{4}\ numbers\ can\ have\ hundreds\ digit. \\ \bold{3}\ numbers\ can\ have\ tens\ digit. \\ \bold{1}\ number\ can\ have\ unit\ digit. \\ \\ \therefore\ 6 \times 5 \times 4 \times 3 \times 1 = \bold{360}\ \ 5-digit\ numbers.$

$\\ \\ On\ 6-digit\ numbers, \\ \\ \bold{6}\ numbers\ can\ have\ lakhs\ digit. \\ \bold{5}\ numbers\ can\ have\ ten\ thousands\ digit. \\ \bold{4}\ numbers\ can\ have\ thousands\ digit. \\ \bold{3}\ numbers\ can\ have\ hundreds\ digit. \\ \bold{2}\ numbers\ can\ have\ tens\ digit. \\ \bold{1}\ number\ can\ have\ unit\ digit. \\ \\ \therefore\ 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \bold{720}\ \ 6-digit\ numbers.$

$\\ \\ On\ 7-digit\ numbers, \\ \\ \bold{6}\ numbers\ can\ have\ ten\ lakhs\ digit. \\ \bold{5}\ numbers\ can\ have\ lakhs\ digit. \\ \bold{4}\ numbers\ can\ have\ ten\ thousands\ digit. \\ \bold{3}\ numbers\ can\ have\ thousands\ digit. \\ \bold{2}\ numbers\ can\ have\ hundreds\ digit. \\ \bold{1}\ number\ can\ have\ tens\ digit. \\ \bold{1}\ number\ (0)\ can\ have\ unit\ digit. \\ \\ \therefore\ 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 1 = \bold{720}\ \ 7-digit\ numbers.$

$\\ \\ 8\ or\ more\ digit\ numbers\ can't\ be\ formed\ as\ repetition\ will\ occur. \\ \\ \therefore\ 6 + 30 + 120 + 360 + 720 + 720 = \bold{1956}\ even\ natural\ numbers \\ which\ is\ divisible\ by\ 5\ can\ be\ formed.$

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