Question

How many different numbers of 7 digit(without repetition of digit)can be formed from the digits

2,1,8,0,9,6,5?How many of them will have 0 in the unit’s place?​

Answer 1

Answer:

$\large \green{ \underline{ \sf \: Solution : - }}$

$\sf \: total \: no. \: of \: ways \: = 6 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$\sf \: = 4320 \: ways$

$\sf \: Alternatively \: 7 ! - 6 ! = 4320 \: ways$

$\sf \: \green{When \: zero \: is \: in \: unit \: place}$

$\sf \: no \: of \: ways \: = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 1 \\ \sf= 720$

$\sf \: because \: zero \: is \: in \: unit \: place \: then \: it \: is \: fixed \: so \: it \: will \: be \: chosen \: by \: 1 \: ways..$

$\sf \: and \: rest \: 6 \: digits \: in \: 6 \: ways$

Answer 2

totalno.ofways=6×6×5×4×3×2×1

\sf \: = 4320 \: ways=4320ways

\sf \: Alternatively \: 7 ! - 6 ! = 4320 \: waysAlternatively7!−6!=4320ways

\sf \: \green{When \: zero \: is \: in \: unit \: place}Whenzeroisinunitplace

\begin{gathered} \sf \: no \: of \: ways \: = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 1 \\ \sf= 720\end{gathered}

noofways=6×5×4×3×2×1×1

=720

\sf \: because \: zero \: is \: in \: unit \: place \: then \: it \: is \: fixed \: so \: it \: will \: be \: chosen \: by \: 1 \: ways..becausezeroisinunitplacethenitisfixedsoitwillbechosenby1ways..

\sf \: and \: rest \: 6 \: digits \: in \: 6 \: waysandrest6digitsin6ways