The number of values of r satisfying ^{69}C_{ | vedclass

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(B) The given equation is $^{69}C_{3r-1} - ^{69}C_{r^2} = ^{69}C_{r^2-1} - ^{69}C_{3r}$.Rearranging the terms,we get: $^{69}C_{3r-1} + ^{69}C_{3r} = ^

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$2$

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(B) The given equation is $^{69}C_{3r-1} - ^{69}C_{r^2} = ^{69}C_{r^2-1} - ^{69}C_{3r}$.Rearranging the terms,we get: $^{69}C_{3r-1} + ^{69}C_{3r} = ^{69}C_{r^2} + ^{69}C_{r^2-1}$.Using the Pascal's identity $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$,the equation becomes: $^{70}C_{3r} = ^{70}C_{r^2}$.This implies either $3r = r^2$ or $3r + r^2 = 70$.Case $1$: $r^2 - 3r = 0 \Rightarrow r(r-3) = 0$,so $r = 0$ or $r = 3$.Case $2$: $r^2 + 3r - 70 = 0 \Rightarrow (r+10)(r-7) = 0$,so $r = -10$ or $r = 7$.Since the lower index of a combination $^{n}C_{k}$ must be a non-negative integer and $0 \le k \le n$,we check the validity of $r$:For $r=0$,$3r-1 = -1$ (not valid).For $r=3$,$3r-1=8, 3r=9, r^2=9, r^2-1=8$ (all valid).For $r=-10$,$3r-1 = -31$ (not valid).For $r=7$,$3r-1=20, 3r=21, r^2=49, r^2-1=48$ (all valid).Thus,the valid values for $r$ are $3$ and $7$.The number of such values is $2$.

The number of values of $r$ satisfying $^{69}C_{3r-1} - ^{69}C_{r^2} = ^{69}C_{r^2-1} - ^{69}C_{3r}$ is:

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