$25$ 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..

Part b should have been $\frac{60 \times C(22,4) \times C(18,4) \times C(14,4) \times C(10,5) \times C(5,5)}{ C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
part c should have been $\frac{20 \times C(22,3) \times C(19,4) \times C(15,5) \times C(10,5) \times C(5,5)}{ C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
