In a collection of six english books and 4 Hindi books(non identical),the probability that three English books are placed together is?
The solution I found first paired those $3$ English books as $1$, so total books now are $8$ and arrangements are $8!$(also arrangements of $3$ paired books is $3!$). And total posible arrangments were $10!$
So $P= 8!3!/10!$
But my doubt is for pairing these $3$ books we first must select $3$ books out of $6$, thus multiplying required Probability by $(6C3)$
Taking the question to mean P(at least $3$ English books are together)
Since probability has been asked for, we can simply use combinations
Also, here it is simpler to use the complement, weeding out invalid combos.
There are a total of $\Large\binom{10}4$ possible combinations
Consider ten English books arranged $\boxed{EE}\; H\;\boxed{ E E}\; H\;\boxed{ E E}\; H\;\boxed{ E E}\; H\;\boxed{ E E}$
All these arrangements are invalid, and will remain invalid in whatever manner we take out $4$ English books to reduce them to the desired $6$
Removal of $4$ English books can be all from two boxes, $\Large\binom52$, one each from four boxes, $\Large\binom54$, or two from one box and one each from another two $\Large\binom51\binom42$
Putting the pieces together, the required probability is
$$\Large 1 - \frac{\binom52 + \binom54 + \binom51\binom42}{\binom{10}4} =\frac{11}{14}$$
If you want to see another approach , you can use complement rule such that $$1 -\text{there is not any three adjacent Eng. book}$$
So , if there is not any three adjacent Eng. book , we can have :
Align $4$ hindi books in a line $4$ ways such that $$-\fbox{H}-\fbox{H}-\fbox{H}-\fbox{H}-$$
We have $5$ empty spaces to place English books. Now , select the place for a pair of English book by $C(5,1)$, the rest of English books will be distributed into the other empty spaces where each empty space will take only one book.Finally , arrange the English books by $6!$. Then , the probability is $$\frac{5 \times6!\times4!}{10!}$$
Select the places for pairs by $C(5,2)=10$ ways , the other $2$ books will be placed the rest places where each places have exactly one English book, so we can select te places for them among $3$ by $C(3,2)=3 $ ways. Then, the probability is $$\frac{10 \times3 \times6!\times4!}{10!}$$
Select $3$ places among $5$ to places the pairs by $C(5,3)=10$ ways , so the probability is $$\frac{10 \times6!\times4!}{10!}$$
By, complement rule : $$1 -\bigg(\frac{5 \times6!\times4!}{10!}+\frac{30 \times6!\times4!}{10!}+\frac{10 \times6!\times4!}{10!}\bigg)$$
$$1- \frac{45 \times6!\times4!}{10!}=\frac{11}{14} =0.785714285$$
Addendum added to (also) answer the alternative question of exactly 3 English books together.
From the posted analysis, I am inferring (perhaps wrongly) that the question is asking for the probability that in a random arrangement of the $(10)$ books, there is some portion of the arrangement where at least $(3)$ English books are together.
My inference is based on the OP's (i.e. original poster's) consideration of $(3)$ of the English books as 1 unit, and then permuting the $(8)$ units. So, the OP makes no mention of preventing the books before or after the $3$-book unit from being English.
The OP's work is good, until the end. He is right that you have to adjust for the fact that there are $~\displaystyle \binom{6}{3} = 20~$ ways of selecting the $(3)$ English books that will become the $3$-book unit.
However, you can not merely multiply the fraction $~\displaystyle \frac{8! \times 3!}{(10)!}~$ by $(20)$.
The reason is because of over-counting. For example, suppose that the books are labeled E-1, E-2, E-3, E-4, E-5, E-6, H-1, H-2, H-3, H-4.
Further suppose that you are considering the specific arrangement of
E-1:E-2:E-3:E-4:E-5:E-6:H-1:H-2:H-3:H-4.
In the computation of $~\displaystyle \frac{8! \times 3!}{(10)!} \times (20),$ the above (single) arrangement is over-counted, because you can construe the arrangement of $(3)$ English books to be any one of
E-1:E-2:E-3, or E-2:E-3:E-4, or E-3:E-4:E-5, or E-4:E-5:E-6.
There are a variety of remedies. Initially, I explored extending the OP's direct approach, but found the Math too convoluted. Then, I considered Stars and Bars theory.
See this article and this article for introductions to the theory.
Since this is a Probability problem, you can ignore the considerations of the order that the English books are permuted or the order that the Hindi books are permuted. That is, you can assume, without loss of generality, that only those arrangements where the English books occur in the order E-1,E-2, ..., E-6 are relevant and that only those arrangements where the Hindi books also occur in the order H-1, ..., H-4 are relevant. That is the order that either the English or Hindi books occur does not affect the probability of $(3)$ or more English books being together.
Instead, you can analyze as follows:
The *'s represent the potential placement of the Hindi books and the =='s represent the placement of the $(6)$ English books.
The number of ways of placing the $(4)$ Hindi books in the $(7)$ regions before/after the English books is equal to the number of non-negative integer solutions to
$$x_1 + x_2 + \cdots + x_7 = 4. $$
From Stars and Bars theory, there are
$$\binom{10}{4} = 210 ~~\text{solutions}. \tag1 $$
Each of these solutions is equally likely to occur. So, the problem reduces to determining how many of these $(210)$ solutions allow $(3)$ or more English books to be together.
The only way that you can have $(3)$ or more English books together is if (ignoring the outer variables $x_1, x_7$) there are two consecutive inner variables that are equal to $(0)$.
I will use Inclusion-Exclusion to enumerate the number of satisfying solutions. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
Using the syntax of the 2nd Inclusion-Exclusion link, let :
$S_1$ denote the subset of solutions where $x_2,x_3$ are both equal to $(0)$.
$S_2$ denote the subset of solutions where $x_3,x_4$ are both equal to $(0)$.
$S_3$ denote the subset of solutions where $x_4,x_5$ are both equal to $(0)$.
$S_4$ denote the subset of solutions where $x_5,x_6$ are both equal to $(0)$.
Then the overall probability will be
$$\frac{|S_1 \cup S_2 \cup S_3 \cup S_4|}{210}.$$
For $k \in \{1,2,3,4\}$, let $T_k$ denote the sum of the $~\displaystyle \binom{4}{k}~$ terms given by
$$\sum_{1 \leq i_1 < \cdots < i_k \leq 4} \left|S_{i_1} \cap \cdots \cap S_{i_k}\right|.$$
Then, in accordance with Inclusion-Exclusion theory, the desired probability will be
$$\frac{T_1 - T_2 + T_3 - T_4}{210}.$$
$\underline{\text{Computation of} ~T_1}$
First, compute $|S_1|$. This means that you are enumerating the number of non-negative solutions to
$$x_1 + 0 + 0 + x_4 + x_5 + x_6 + x_7 = 4 ~: ~\binom{8}{4} = 70 ~~\text{solutions}.$$
By symmetry, $|S_1| = |S_2| = |S_3| = |S_4|.$
Therefore,
$$T_1 = 4 \times 70 = 280. \tag2 $$
$\underline{\text{Computation of} ~T_2}$
First, compute $|S_1 \cap S_2|$. This means that you are enumerating the number of non-negative solutions to
$$x_1 + 0 + 0 + 0 + x_5 + x_6 + x_7 = 4 ~: ~\binom{7}{3} = 35 ~~\text{solutions}.$$
By symmetry, $|S_1 \cap S_2| = |S_2 \cap S_3| = |S_3 \cap S_4|.$
Now, compute $|S_1 \cap S_3|$. This means that you are enumerating the number of non-negative solutions to
$$x_1 + 0 + 0 + 0 + 0 + x_6 + x_7 = 4 ~: ~\binom{6}{2} = 15 ~~\text{solutions}.$$
By symmetry, $|S_1 \cap S_3| = |S_1 \cap S_4| = |S_2 \cap S_4|.$
Therefore,
$$T_2 = \left(3 \times 35\right) + \left(3 \times 15\right) = 150. \tag3 $$
$\underline{\text{Computation of} ~T_3}$
First, compute $|S_1 \cap S_2 \cap S_3|$. This means that you are enumerating the number of non-negative solutions to
$$x_1 + 0 + 0 + 0 + 0 + x_6 + x_7 = 4 ~: ~\binom{6}{2} = 15 ~~\text{solutions}.$$
By symmetry, $|S_1 \cap S_2 \cap S_3| = |S_2 \cap S_3 \cap S_4|.$
Now, compute $|S_1 \cap S_2 \cap S_4|$. This means that you are enumerating the number of non-negative solutions to
$$x_1 + 0 + 0 + 0 + 0 + 0 + x_7 = 4 ~: ~\binom{5}{1} = 5 ~~\text{solutions}.$$
By symmetry, $|S_1 \cap S_2 \cap S_4| = |S_1 \cap S_3 \cap S_4|.$
Therefore,
$$T_3 = \left(2 \times 15\right) + \left(2 \times 5\right) = 40. \tag4 $$
$\underline{\text{Computation of} ~T_4}$
Compute $|S_1 \cap S_2 \cap S_3 \cap S_4|$. This means that you are enumerating the number of non-negative solutions to
$$x_1 + 0 + 0 + 0 + 0 + 0 + x_7 = 4 ~: ~\binom{5}{1} = 5 ~~\text{solutions}.$$
Therefore,
$$T_4 = 5. \tag5 $$
$\underline{\text{Final Computation}}$
$$\frac{(T_1 + T_3) - (T_2 + T_4)}{210} = \frac{(280 + 40) - (150 + 5)}{210} = \frac{165}{210} = \frac{11}{14}.$$
Addendum
Addendum added to (also) answer the alternative question of exactly 3 English books together. This answer assumes that the situation where there are $(2)$ groups of 3 consecutive English books each is also regarded as satisfactory.
The easiest way to answer this question is to use the analysis in the first part of my answer. Then, I can similarly enumerate the number of ways (out of $210$) of having at least $(4)$ consecutive English books. Then, I can deduct this computation from $(165)$.
As before, consider the equation in non-negative integers
$$x_1 + x_2 + \cdots + x_7 = 4.$$
The only way of having at least $(4)$ consecutive English books is if there are $(3)$ consecutive inner variables (i.e. $x_2, \cdots, x_6$) that are equal to $(0)$.
$S_1$ denote the subset of solutions where $x_2,x_3,x_4$ are all equal to $(0)$.
$S_2$ denote the subset of solutions where $x_3,x_4,x_5$ are all equal to $(0)$.
$S_3$ denote the subset of solutions where $x_4,x_5,x_6$ are all equal to $(0)$.
For $k \in \{1,2,3\}$, let $T_k$ denote the sum of the $~\displaystyle \binom{3}{k}~$ terms given by
$$\sum_{1 \leq i_1 < \cdots < i_k \leq 4} \left|S_{i_1} \cap \cdots \cap S_{i_k}\right|.$$
Then, the final computation will be
$$\frac{165 - [T_1 - T_2 + T_3]}{210}.$$
$\underline{\text{Computation of} ~T_1}$
First, compute $|S_1|$. This means that you are enumerating the number of non-negative solutions to
$$x_1 + 0 + 0 + 0 + x_5 + x_6 + x_7 = 4 ~: ~\binom{7}{4} = 35 ~~\text{solutions}.$$
By symmetry, $|S_1| = |S_2| = |S_3|.$
Therefore,
$$T_1 = 3 \times 35 = 105. $$
$\underline{\text{Computation of} ~T_2}$
First, compute $|S_1 \cap S_2|$. This means that you are enumerating the number of non-negative solutions to
$$x_1 + 0 + 0 + 0 + 0 + x_6 + x_7 = 4 ~: ~\binom{6}{2} = 15 ~~\text{solutions}.$$
By symmetry, $|S_1 \cap S_2| = |S_2 \cap S_3|.$
Now, compute $|S_1 \cap S_3|$. This means that you are enumerating the number of non-negative solutions to
$$x_1 + 0 + 0 + 0 + 0 + 0 + x_7 = 4 ~: ~\binom{5}{1} = 5 ~~\text{solutions}.$$
Therefore,
$$T_2 = \left(2 \times 15\right) + 5 = 35. \tag3 $$
$\underline{\text{Computation of} ~T_3}$
Compute $|S_1 \cap S_2 \cap S_3|$. This means that you are enumerating the number of non-negative solutions to
$$x_1 + 0 + 0 + 0 + 0 + 0 + x_7 = 4 ~: ~\binom{5}{1} = 5 ~~\text{solutions}.$$
Therefore,
$$T_4 = 5. $$
$\underline{\text{Final Computation}}$
$$\frac{(165 + T_2) - (T_1 + T_3)}{210} = \frac{(165 + 35) - (105 + 5)}{210} = \frac{90}{210} = \frac{3}{7}.$$
