help

[fathom knoll]

Work out how many 5-digit odd numbers can be
formed using the digits 1 3 4 6 8 with no
repetition of any digit

<@&791435371564892232>

[versed crystal]

how many possibilities for 1st digit

after that how many for 2nd (1 digit is used and cant be used again

then multiply

you will find factorial is useful

[fathom knoll]

5 I think

[versed crystal]

yes

[fathom knoll]

5! then?

but I don't think u could do that

[versed crystal]

lmao why not

[fathom knoll]

bc it says odd number

[versed crystal]

oh lmao

[fathom knoll]

so the last digit has to be odd

yh cos I thought that in the beginning asw

but then I realised

[versed crystal]

so the last digit cant be even

[fathom knoll]

yh

[versed crystal]

find every case where is end in 4,6 8

if 4 is reserved for last digit then it becomes 4*3*2*1

same for 6 and 8

and subtract from total possibilities

[fathom knoll]

wait why are we ending in 4 6 or 8 is it not meant to be even

ohh

ok

yh subtract

[versed crystal]

so im doing total - everything we dont want

[fathom knoll]

got u

[versed crystal]

yes

[fathom knoll]

could u not js do the same thing but with 1 and 3 instead

so instead of subtracting

js do 4! + 4!

which is 48

[fathom knoll]

and it gives u same answer as this

[versed crystal]

u could im pretty sure

😲

[fringe heart]

realistically

wouldnt you do the whole 5x4x3x2 everything like you said

then the 4x3x2x1 or what