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Question-214595




Question Number 214595 by malwan last updated on 13/Dec/24
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Commented by malwan last updated on 13/Dec/24
I send jpg photo
$${I}\:{send}\:{jpg}\:{photo} \\ $$
Commented by malwan last updated on 13/Dec/24
but I cant see it here
$${but}\:{I}\:{cant}\:{see}\:{it}\:{here} \\ $$
Commented by Tinku Tara last updated on 13/Dec/24
Ok. Will check
$$\mathrm{Ok}.\:\mathrm{Will}\:\mathrm{check} \\ $$
Commented by malwan last updated on 13/Dec/24
Commented by malwan last updated on 13/Dec/24
this is my question
$${this}\:{is}\:{my}\:{question} \\ $$
Commented by mr W last updated on 13/Dec/24
i got  2×3!×(4!−2×3!)=144 ways
$${i}\:{got} \\ $$$$\mathrm{2}×\mathrm{3}!×\left(\mathrm{4}!−\mathrm{2}×\mathrm{3}!\right)=\mathrm{144}\:{ways} \\ $$
Commented by malwan last updated on 13/Dec/24
Can you explane it mr W ,  please  ⋛
$${Can}\:{you}\:{explane}\:{it}\:{mr}\:{W}\:, \\ $$$${please}\:\:\cancel{\lesseqgtr} \\ $$
Commented by mr W last updated on 13/Dec/24
since it is not a round table, we  should define some rules. e.g.  following two arrangements are  seen as identical.
$${since}\:{it}\:{is}\:{not}\:{a}\:{round}\:{table},\:{we} \\ $$$${should}\:{define}\:{some}\:{rules}.\:{e}.{g}. \\ $$$${following}\:{two}\:{arrangements}\:{are} \\ $$$${seen}\:{as}\:{identical}. \\ $$
Commented by mr W last updated on 13/Dec/24
Commented by mr W last updated on 13/Dec/24
but following two arrangements  are seen as different.
$${but}\:{following}\:{two}\:{arrangements} \\ $$$${are}\:{seen}\:{as}\:{different}. \\ $$
Commented by mr W last updated on 13/Dec/24
Commented by mr W last updated on 13/Dec/24
now we can try to solve.  we have 4 even numbers. they may  not be placed adjacently to each  other. i.e. they should all on the  corners or all in the middle of each  side. there are 2 possibilities. to  place the 4 even numbers around  the table there are 3! ways. now  we can place the 4 odd numvers  in the 4 brackets between the even  numbers. there are 4! ways, but the  number 3 may not be placed on the  left or right side of the number 6,  so the number of valid ways is  4!−2×3!.   totally there are  2×3!×(4!−2×3!)=144 ways.
$${now}\:{we}\:{can}\:{try}\:{to}\:{solve}. \\ $$$${we}\:{have}\:\mathrm{4}\:{even}\:{numbers}.\:{they}\:{may} \\ $$$${not}\:{be}\:{placed}\:{adjacently}\:{to}\:{each} \\ $$$${other}.\:{i}.{e}.\:{they}\:{should}\:{all}\:{on}\:{the} \\ $$$${corners}\:{or}\:{all}\:{in}\:{the}\:{middle}\:{of}\:{each} \\ $$$${side}.\:{there}\:{are}\:\mathrm{2}\:{possibilities}.\:{to} \\ $$$${place}\:{the}\:\mathrm{4}\:{even}\:{numbers}\:{around} \\ $$$${the}\:{table}\:{there}\:{are}\:\mathrm{3}!\:{ways}.\:{now} \\ $$$${we}\:{can}\:{place}\:{the}\:\mathrm{4}\:{odd}\:{numvers} \\ $$$${in}\:{the}\:\mathrm{4}\:{brackets}\:{between}\:{the}\:{even} \\ $$$${numbers}.\:{there}\:{are}\:\mathrm{4}!\:{ways},\:{but}\:{the} \\ $$$${number}\:\mathrm{3}\:{may}\:{not}\:{be}\:{placed}\:{on}\:{the} \\ $$$${left}\:{or}\:{right}\:{side}\:{of}\:{the}\:{number}\:\mathrm{6}, \\ $$$${so}\:{the}\:{number}\:{of}\:{valid}\:{ways}\:{is} \\ $$$$\mathrm{4}!−\mathrm{2}×\mathrm{3}!.\: \\ $$$${totally}\:{there}\:{are} \\ $$$$\mathrm{2}×\mathrm{3}!×\left(\mathrm{4}!−\mathrm{2}×\mathrm{3}!\right)=\mathrm{144}\:{ways}. \\ $$
Commented by malwan last updated on 13/Dec/24
thank you so much sir   ⋛
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$$$\:\lesseqgtr \\ $$

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