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Question Number 124916 by liberty last updated on 07/Dec/20

The number of ways arrangements of the word ′MASKARA′ with exactly 2 A′s are adjacent??

$${The}\:{number}\:{of}\:{ways}\:{arrangements}\: \\ $$$${of}\:{the}\:{word}\:'{MASKARA}'\:{with}\:{exactly} \\ $$$$\mathrm{2}\:{A}'{s}\:\:{are}\:{adjacent}??\: \\ $$

Answered by mr W last updated on 07/Dec/20

_M_S_K_R_ 4!×P_2 ^5 =480

$$\_\mathrm{M\_S\_K\_R\_} \\ $$$$\mathrm{4}!×{P}_{\mathrm{2}} ^{\mathrm{5}} =\mathrm{480} \\ $$

Commented by mr W last updated on 07/Dec/20

Method II 3 A′s adjacent: 5! at least 2 A′s adjacent: 6! exactly 2 A′s adjacent: 6!−2×5!=480

$${Method}\:{II} \\ $$$$\mathrm{3}\:{A}'{s}\:{adjacent}:\:\mathrm{5}! \\ $$$${at}\:{least}\:\mathrm{2}\:{A}'{s}\:{adjacent}:\:\mathrm{6}! \\ $$$${exactly}\:\mathrm{2}\:{A}'{s}\:{adjacent}:\:\mathrm{6}!−\mathrm{2}×\mathrm{5}!=\mathrm{480} \\ $$

Commented by bemath last updated on 07/Dec/20

i got 960 sir. what wrong?

$${i}\:{got}\:\mathrm{960}\:{sir}.\:{what}\:{wrong}? \\ $$

Commented by mr W last updated on 07/Dec/20

how did you get?

$${how}\:{did}\:{you}\:{get}? \\ $$

Answered by liberty last updated on 07/Dec/20

(•) −_1^(st) −^(AA) −_2^(nd) −^A −_3^(rd) Treat 4 letters M ,S, K, R as identical ′x′ since exactly 2 A′s are adjacent, one x must be put in 2^(nd) places ⇒ −_1^(st) AA −_2^(nd) ^x A −_3^(rx) the remaining 3 x′s can be put in (((3+3−1)),(( 3)) ) = ((5),(3) ) = 10 (••) −_1^(st) A −_2^(nd) AA −_3^(rd) similar to cases (•) therefore the desired number of ways is given by 2× ((5),(3) )×4! = 20×24 = 480

$$\left(\bullet\right)\:\underset{\mathrm{1}^{{st}} } {−}\:\overset{{AA}} {−}\:\underset{\mathrm{2}^{{nd}} } {−}\:\overset{{A}} {−}\:\underset{\mathrm{3}^{{rd}} } {−} \\ $$$$\:{Treat}\:\mathrm{4}\:{letters}\:{M}\:,{S},\:{K},\:{R}\:{as}\:{identical}\:'{x}' \\ $$$${since}\:{exactly}\:\mathrm{2}\:{A}'{s}\:{are}\:{adjacent},\:{one}\:{x}\:{must}\:{be} \\ $$$${put}\:{in}\:\mathrm{2}^{{nd}} \:{places}\:\Rightarrow\:\underset{\mathrm{1}^{{st}} } {−}\:{AA}\:\underset{\mathrm{2}^{{nd}} } {\overset{{x}} {−}}\:{A}\:\underset{\mathrm{3}^{{rx}} } {−} \\ $$$${the}\:{remaining}\:\mathrm{3}\:{x}'{s}\:{can}\:{be}\:{put}\:{in}\:\begin{pmatrix}{\mathrm{3}+\mathrm{3}−\mathrm{1}}\\{\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{5}}\\{\mathrm{3}}\end{pmatrix}\:=\:\mathrm{10} \\ $$$$\left(\bullet\bullet\right)\:\underset{\mathrm{1}^{{st}} } {−}\:{A}\:\underset{\mathrm{2}^{{nd}} } {−}\:{AA}\:\underset{\mathrm{3}^{{rd}} } {−}\:{similar}\:{to}\:{cases}\:\left(\bullet\right) \\ $$$${therefore}\:{the}\:{desired}\:{number}\:{of}\:{ways} \\ $$$${is}\:{given}\:{by}\:\mathrm{2}×\begin{pmatrix}{\mathrm{5}}\\{\mathrm{3}}\end{pmatrix}×\mathrm{4}!\:=\:\mathrm{20}×\mathrm{24}\:=\:\mathrm{480} \\ $$

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