Let m (respectively,n) be the number of 5-dig | vedclass

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(C) For the sum of two adjacent digits to be odd,one must be even and the other must be odd. The set of digits is $\{1, 2, 3, 4, 5\}$,where odd digits

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) For the sum of two adjacent digits to be odd,one must be even and the other must be odd. The set of digits is $\{1, 2, 3, 4, 5\}$,where odd digits are $O = \{1, 3, 5\}$ (count $3$) and even digits are $E = \{2, 4\}$ (count $2$).Case $I$: With repetition $(m)$Pattern $1$: $O-E-O-E-O$. Number of ways $= 3 	imes 2 	imes 3 	imes 2 	imes 3 = 108$.Pattern $2$: $E-O-E-O-E$. Number of ways $= 2 	imes 3 	imes 2 	imes 3 	imes 2 = 72$.Total $m = 108 + 72 = 180$.Case $II$: Without repetition $(n)$Pattern $1$: $O-E-O-E-O$. Number of ways $= 3 	imes 2 	imes 2 	imes 1 	imes 1 = 12$.Pattern $2$: $E-O-E-O-E$. This is impossible because we only have $2$ even digits,and this pattern requires $3$ even digits.Total $n = 12$.Therefore,$\frac{m}{n} = \frac{180}{12} = 15$.

Let $m$ (respectively,$n$) be the number of $5$-digit integers obtained by using the digits $1, 2, 3, 4, 5$ with repetitions (respectively,without repetitions) such that the sum of any two adjacent digits is odd. Then $\frac{m}{n}$ is equal to

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