3

Methods 1 and 2 should work. ${16\choose4}-{8\choose4}=1750$ ${8\choose1}{8\choose3}+{8\choose2}{8\choose2}+{8\choose3}{8\choose1}+{8\choose4}=1750$ Method 3 does not correctly count the set. You would be counting M1 + M2W1W2 as a different arrangement than M2 + M1W1W2 even though they represent the same team.


3

The "two numbers chosen at random" indicates that two numbers are going to be chosen randomly from $1,2,...9$ by a lottery. You have tickets numbered $1$ and $2$, so if either of those come up in the chosen numbers, you win one prize. If the two chosen numbers are $1$ and $2$, you win both the prizes. Hope its clear enough now.


3

I think your solution is optimal. However, if you really want to see a combinatorial solution, keep reading. There are $\binom{9}{4}$ to select the other four members of the group containing person $A$. If person $A$ and person $B$ are in the same group, there are $\binom{8}{3}$ ways to select the other three members of their group. Hence, the probability ...


3

When you remove the 'single color' strominoes (e.g. $000, 111$), you also need to remove the symmetric $010,020$, etc.., as these are also doubled up. Your calculation then becomes: $$\frac{(7^3-7^2)}{2}+7^2=\frac{(343-49)}{2}+7^2=\frac{294}{2}+49=147+49=196$$


2

Because there are fewer orders if some of the elements match. If you have the combination $1,1,2$ there are only $3$, not $3!=6$ orders for these elements. You therefore cannot just multiply the number of combinations by $k!$ to get the number of permutations.


2

Proof by induction. When $n=0$ then $\mathcal P(\emptyset)=\{\emptyset\}$ so $|\mathcal P(\emptyset)|=1.$ Now, assume it is true for all $|S|=n,$ and assume $|T|=n+1.$ Pick one element $x\in T.$ Then let $T_0=T\setminus \{x\}.$ We have $|T_0|=n.$ Then for every $M\subseteq T_0$ there are exactly two subsets of $T,$ $M$ and $M\cup\{x\}.$ So $$|\mathcal P(T)|=...


1

Well, if the exact wording is "without repetitions" I would interpret it as no letter being used more than once in the word, which makes it quite easy. There are $9$ distinct letters, $I\,N\,E\,Q\,U\,A\,L\,T\,S$ thus number of number of $3$ letter words = $P^9_3 = 9\cdot8\cdot7$


1

Addendum added to respond to OP's comment. I agree with coffeemath's comments. Your $\displaystyle \frac{7!}{4!}$ computation represents 3 letter words that exclude any "I" or "E" letters. I'm unsure what your $\displaystyle\binom{5}{2}$ computation is then supposed to represent. Personally, I would break the situation into mutually ...


1

If there were no men on the table, the women could be permuted in $4!/4=3!$ ways. But once the men's seats are fixed, each of the $4$ seating arrangements of the women which were earlier treated as equivalent have to now be treated as distinct because of the changed relative position of men and women. This graph may help you understand better: In both ...


1

You have 4 coin tosses, which can be thought of as 4 ordered slots (coin values) to fill with either heads or tails. In this case, out of 4 possible slots, you are selecting 3 slots without replacement. So, it's not the coin values themselves, but the slots into which you want the heads to fall into. The chosen slots will contain heads, and the remaining ...


1

Suppose $S=\{1,2,3,4,5,6,7,8\}$, and you are choosing $5$ of those, but $A=\{2,3\}$ and $B=\{4,5,6\}$ are mutually exclusive subsets (you can only have at most $1$ from each.) In this example it so happens to be barely possible to satisfy the requirements. You must take exactly one of $\{2,3\}$ and one of ${4,5,6}$ and all of $C = \{1,7,8\}$ so there are $2\...


1

Comment ${32 \choose 2}=\frac{31\times 32}2$ ${32 \choose 3}=5\times 31\times 32=2\times 64\times 31+2\times\frac{31\times 32}2$ $\Rightarrow 2\times 1984+2{32 \choose 2}={32 \choose 3}\rightarrow 1984=\frac12 [{32 \choose 3}+{32 \choose 2}]-\frac32{32 \choose 2}$ Using formula ${n+1 \choose r}={n \choose r}+{n \choose r-1}$, we obtain: $1984=\frac12{33 \...


1

The reason that there are $2^n$ subsets of an $n$-element set is that we can choose to include $0$ or $1$ of the first element, $0$ or $1$ of the second, and so on. If we have a multiset with $n_k$ copies of item $k$, $k=1,2,\dots,m$ then we can include from $0$ to $n_1$ copies of the first item, from $0$ to $n_2$ copies of the second, and so on, so the ...


1

We can think like there are two groups such as group $A$ and group $B$ Lets say that they both are in group $A$ ,then the the probability of being $A$ can be calculated $\frac{C(8,3) }{C(10,5)} \times \frac {C(5,5) }{C(5,5)}=\frac{2}{9}$ .However , they might have been in group $B$ , so we should multiply by $2$ .Hence , the result is $\frac{4}{9}$


1

Hint: Number of handshakes of american woman with each other would be $^5C_2=10$. Absolutely. However second time you chose $^{15}C_2=10$ you have again took the number of second american woman so you need to subtract $^{15}C_2-^5C_2$


1

The problem is that you have counted handshakes between pairs of American women twice: once in the handshakes not involving men and once in the handshakes not involving Indian women. There are $^5C_2=10$ of these, which need to be subtracted.


1

We can even make a stronger statement that $${n \choose k} = \Theta\left (\left( \frac{n}{k} \right)^k \right)$$ For the proof, see this.


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