$T_{1}$ |
$T_{2}$ |
$T_{3}$ |
$T_{4}$ |
$1.$ |
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$2.$ |
$r(x)$ |
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$3.$ |
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$4.$ |
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$w(x)$ |
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$5.$ |
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$6.$ |
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$Commit$ |
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$7.$ |
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$8.$ $w(x)$ |
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$9.$ |
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$10.$ $Commit$ |
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$11.$ |
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$12.$ |
$r(y)$ |
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$13.$ |
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$14.$ |
$w(z)$ |
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$15.$ |
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$16.$ |
$Commit$ |
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$17.$ |
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$18.$ |
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$r(y)$ |
$19.$ |
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$20.$ |
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$Commit$ |
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Now, See for $T_{2}$ $r(y)$ , $w(z)$ , $Commit$ has no conflict with $T_{1}$ or $T_{3}$. So, $T_{2}$ can be placed in $3,5,7,9,11$ th places. Now all 3 of them can be placed in 1 place , or 2 place in one among these 5 places and 3rd one in another place , or all 3 can place in different place
So, $\binom{5}{1}+\binom{5}{2}+\binom{5}{3}=25$ arrangement are possible
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Now, Similarly $T_{4}$ can arrange in 8 places. Now, $T_{4}$ transaction can place $\binom{8}{1}+\binom{8}{2}=36$ ways
So, total arrangement $25+36=61$ ways