$25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that $25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that
a-) Dennis , John and  Jack are in the same group.
b-)Dennis , John and  Jack are in different groups.
c-)Dennis and John are in the same group but not Jack
My attempt :
a-) If they are in same group , then there are $5$ ways to select this group. Moreover , we should select $2$ people for the group by $C(22,2)$ . Then $5 \times \frac{C(22,2) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
b-)If they are in different groups , we can disribute them by $P(5,3)=60$ ways.Then , $60 \times \frac{C(24,4) \times C(19,4) \times C(14,4) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
c-)We can choose $5$ groups for Dennis and John , 4 groups for Jack ,so $P(5,2)=20$ ways. Then ; $20 \times \frac{C(23,3) \times C(19,4) \times C(15,5) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
Is my solution way correct ? If not ,can you help..

While there is a simpler method as given in one of the answers, even going by your method the working can be simplified quite a bit.
a) Once you fix Dennis's group, there are $4$ more slots in the group to be picked from $24$ people. But if John and Jack are in the same group, you are left with choosing two spots from $22$ people. So desired probability is
$\displaystyle \small {22 \choose 2} / {24 \choose 4} = \frac{1}{46}$
b) The number of arrangements with Dennis, Jack and John being in different groups
$ = \displaystyle {5 \choose 3} \cdot 3! \cdot \frac{22!}{ (5!)^2 (4!)^3}$
Unrestricted arrangements $ = \displaystyle \frac{25!}{ (5!)^5}$
So dividing, you get desired probability of $\displaystyle \frac{25}{46}$
c) Here I will just use result from a) and b). If we add results from a) and b) and subtract from $1$, that gives us probability that any two of them are together while the third is in a different group. But there are $3$ ways to choose which two of them are together and it is equally likely that any two of them are together. So probability that Dennis and John are in the same group but not Jack will be
$ \displaystyle \small \frac{1}{3} \cdot \big(1 - \frac{1}{46} - \frac{25}{46}\big) = \frac{10}{69}$
