From a month of 31 days,3 different dates are | vedclass

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(NONE) Total ways to select $3$ distinct dates from $31$ is $\binom{31}{3} = \frac{31 	imes 30 	imes 29}{3 	imes 2 	imes 1} = 4495$.Let the dates

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(NONE) Total ways to select $3$ distinct dates from $31$ is $\binom{31}{3} = \frac{31 	imes 30 	imes 29}{3 	imes 2 	imes 1} = 4495$.Let the dates be $d_1, d_2, d_3$ such that $1 \le d_1 The conditions are $d_1 \ge 1$ and $d_3 \le 31$.Substituting,$a-d \ge 1 \implies a \ge d+1$ and $a+d \le 31 \implies a \le 31-d$.For a fixed $d$,the number of possible values for $a$ is $(31-d) - (d+1) + 1 = 31-2d$.Since $a$ must exist,$31-2d \ge 1 \implies 2d \le 30 \implies d \le 15$.The total number of such $A$.$P$.s is $\sum_{d=1}^{15} (31-2d) = 29 + 27 + 25 + ... + 1$.This is an arithmetic series with $15$ terms,sum $= \frac{15}{2}(29+1) = 15 	imes 15 = 225$.Probability $= \frac{225}{4495} = \frac{45}{899}$.Here $a = 45, b = 899$. Since $	ext{gcd}(45, 899) = 1$,$a+b = 45 + 899 = 944$.

From a month of $31$ days,$3$ different dates are selected at random. If the probability that these dates are in an increasing $A$.$P$. is equal to $\frac{a}{b}$,where $a, b \in N$ and $\text{gcd}(a, b) = 1$,then $a + b$ is equal to ————

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