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Question Number 124829 by bramlexs22 last updated on 06/Dec/20

How many ways are there to arrange the letters of the word ′ALAMATAR′ if no two A′s are adjacent?

$$\:{How}\:{many}\:{ways}\:{are}\:{there}\:{to}\:{arrange} \\ $$$${the}\:{letters}\:{of}\:{the}\:{word}\:'{ALAMATAR}'\:{if} \\ $$$${no}\:{two}\:{A}'{s}\:{are}\:{adjacent}?\: \\ $$

Answered by mr W last updated on 06/Dec/20

to arrange at first the four letters LMTR, there are 4! ways. □L□M□T□R□ to arrange the four letters A in the five □ positions, there are C_4 ^5 ways. ⇒totally 4!×C_4 ^5 =120

$${to}\:{arrange}\:{at}\:{first}\:{the}\:{four}\:{letters} \\ $$$${LMTR},\:{there}\:{are}\:\mathrm{4}!\:{ways}. \\ $$$$\Box{L}\Box{M}\Box{T}\Box{R}\Box \\ $$$${to}\:{arrange}\:{the}\:{four}\:{letters}\:{A}\:{in}\:{the} \\ $$$${five}\:\Box\:{positions},\:{there}\:{are}\:{C}_{\mathrm{4}} ^{\mathrm{5}} \:{ways}. \\ $$$$\Rightarrow{totally}\:\mathrm{4}!×{C}_{\mathrm{4}} ^{\mathrm{5}} =\mathrm{120} \\ $$

Commented by bramlexs22 last updated on 06/Dec/20

thank you

$${thank}\:{you} \\ $$

Answered by john_santu last updated on 06/Dec/20

(•) We first arrange the 4 A′s in a row one way:≡ −_1^(st) A −_2^(nd) A −_3^(rd) A −_4^(th) A −_5^(th) The treat 4 letters ′L,M,T,R ′ as identical ′x′ . Since no two A′s are adjacent one ′x, must be put in the 2^(nd) , 3^(rd) and 4^(th) places (this can be done in one way)⇒−_1^(st) A −_2^(nd) ^x A −_3^(rd) ^x A −_4^(th) ^x A −_5^(th) Now the remaining 1 ′x′ can be put in the 5 places arbitrarily in (((1+5−1)),(( 1)) ) = ((5),(1) ) = 5 Thus the desired number of ways is given by 5×4! = 5×24 = 120

$$\left(\bullet\right)\:{We}\:{first}\:{arrange}\:{the}\:\mathrm{4}\:{A}'{s}\:{in}\:{a}\:{row}\: \\ $$$${one}\:{way}:\equiv\:\underset{\mathrm{1}^{{st}} } {−}\:{A}\:\underset{\mathrm{2}^{{nd}} } {−}\:{A}\:\underset{\mathrm{3}^{{rd}} } {−}\:{A}\:\underset{\mathrm{4}^{{th}} } {−}\:{A}\:\underset{\mathrm{5}^{{th}} } {−} \\ $$$${The}\:{treat}\:\mathrm{4}\:{letters}\:'{L},{M},{T},{R}\:'\:\:{as}\:{identical}\: \\ $$$$'{x}'\:.\:{Since}\:{no}\:{two}\:{A}'{s}\:{are}\:{adjacent}\:{one}\: \\ $$$$'{x},\:{must}\:{be}\:{put}\:{in}\:{the}\:\mathrm{2}^{{nd}} \:,\:\mathrm{3}^{{rd}} \:{and}\:\mathrm{4}^{{th}} \:{places} \\ $$$$\left({this}\:{can}\:{be}\:{done}\:{in}\:{one}\:{way}\right)\Rightarrow\underset{\mathrm{1}^{{st}} } {−}\:{A}\:\underset{\mathrm{2}^{{nd}} } {\overset{{x}} {−}}\:{A}\:\underset{\mathrm{3}^{{rd}} } {\overset{{x}} {−}}\:{A}\:\underset{\mathrm{4}^{{th}} } {\overset{{x}} {−}}\:{A}\:\underset{\mathrm{5}^{{th}} } {−} \\ $$$${Now}\:{the}\:{remaining}\:\mathrm{1}\:'{x}'\:{can}\:{be}\:{put}\:{in}\:{the} \\ $$$$\mathrm{5}\:{places}\:{arbitrarily}\:{in}\:\begin{pmatrix}{\mathrm{1}+\mathrm{5}−\mathrm{1}}\\{\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{5}}\\{\mathrm{1}}\end{pmatrix}\:=\:\mathrm{5} \\ $$$${Thus}\:{the}\:{desired}\:{number}\:{of}\:{ways}\:{is}\:{given} \\ $$$${by}\:\mathrm{5}×\mathrm{4}!\:=\:\mathrm{5}×\mathrm{24}\:=\:\mathrm{120} \\ $$$$ \\ $$

Commented by bramlexs22 last updated on 06/Dec/20

thank you

$${thank}\:{you} \\ $$

Commented by liberty last updated on 06/Dec/20

it's a good explanation to students so they can easily understand it

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