$25$ 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..
 A: I checked the first one, it is correct, but there is a simpler way to compute the probabilities
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A The first friend can be put in any slot, and the remaining two must be put in the same group, thus
$Pr = 1\cdot \frac4{24}\frac3{23} = \frac1{46}$
B Similarly, $Pr = 1\cdot\frac {20}{24}\cdot\frac{15}{23} = \frac{25}{46}$
C John can be anywhere, the other two have to fill two slots in some other group, thus
$Pr = 1\cdot\frac {20}{24}\frac{4}{23} = \frac{10}{69}$
This avoids long expressions and a lot of cancellations.
PS Also avoids errors, which I have found in your working after checking B and C
A: Your solution is correct, as you're considering all equiprobable distributions. However, you could make it simpler considering the people instead of the distributions.
a) If Dennis is in a given group, there's 4/24 probability that John belongs to the same group (among 24 other people, 4 are in the same group, 20 in another one and assumably they're independently assigned to a group). Given that, by the same argument, there's 3/23 probability that Jack belongs to the same group as Dennis and John. The probability of this event is therefore 3*4/(23*24) = 1/46.
b) If Dennis is in a given group, there's 20/24 probability that John belongs to another group, and given that a 15/23 probability that Jack belongs to neither the group of Dennis nor the group of John. The probability is 25/46.
c) I let you find a similar reasoning to this one.
A: 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 $\frac{20 \times C(22,3) \times C(19,4) \times C(15,5) \times C(10,5) \times C(5,5)}{ C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
A: 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}$
