Your local grocery store just received a large shipment of apples, oranges, pears, and bananas---there are only 5 of each fruit.  You are shopping at the store and will purchase your fruit for the week.  How many ways can you select $10$ pieces of fruit from your store's supply of apples, oranges, pears, and bananas, if you need at least $2$ oranges and $1$ apple?  I understand that there are 17 spots to choose from because 3 have been chosen and there are a total of 20 fruits. Would I use the combination by repetition formula? I can't quite wrap my head around what to do. Any help would be appreciated.
121945
$$x_1 + x_2 + x_3 + x_4 = 10$$ But $x_1 \ge 1$ and $x_2 \ge2$.  Finally, $x_1&lt;6,  x_2&lt;6, x_3&lt;6$ and $x_4&lt;6$.  I started off by finding the total amount that you can get for the original equation. That would be $C (13, 3)$. Then i calculated for the cases of numbers being greater than or equal to 5. I dont know how to incorporate $x_1\ge1$ and $x_2\ge2$. Any help?  $x_1, x_2, x_3, x_4$ are all integers that could be >= 0(But $x_1$ and $x_2$ have their own restriction).
114550
I was assigned the following problem for homework and would like to double check that I have approached it correctly.  Let $k \geq 2n$. In how many ways can we distribute $k$ sweets to $n$ children, if each child is supposed to get at least 2 of them?  This is a standard stars-and-bars problem, with the added constraint that each child gets at least two. In other words:  $x_1 + x_2 + \cdots + x_n = k$, such that $x_i \geq 2$, for $0 \leq i \leq n$  We can reduce this problem to solving over $y_i$ such that $y_i = x_i - 2$ and $y_i \geq 0$.  $\sum_{i=0}^n y_i = k$ or $\sum_{i=0}^n (x_i - 2) = k$ which reduces to  $\sum_{i=0}^n x_i = k + 2n$.  Plugging this into standard stars and bars we get $\binom{n + k + 2n - 1}{n - 1} = \binom{k + 3n - 1}{n - 1}$  Is this solution correct? Am I far off? This seems weird that the search space is bigger in the constrained version than the regular ($\geq 0$) version.
93524
I got a perfect orange! I heard I can plant this to grow 3 more perfect oranges, but if I shake the tree it might die. How does this work exactly? Does each perfect tree only grow 3 fruits, does it grow 3 fruits every day until it dies or what?  Can I grow perfect fruits only of my native fruit? I planted a perfect of a friend's fruit but it didn't grow. I'm unsure if that was bad luck or if they won't actually grow.
94755
Suppose that I am buying cakes for a party. There are $k$ different types and I intend to buy a total of $n$ cakes. How many different combinations of cakes could I possibly bring to the party?
59404
We have 4 green apples and 60 red apples. Each time we pick one out without replacement. Then what is the expected number of red apples left when all 4 green apples are picked?
36179
I am reviewing take aways. I am having trouble  How do I do $342 - 58$?  For the ones column I made the $2$ into a $12$ so I can do $12 - 8 = 4$ but I must take away one tens. So I get $3 - 5$ in tens column but I cant do $3 - 5$. What do I do now do I borrow something else?
2129
Started with an algorithms problem which says item is sold in 3 different sizes of boxes. These 3 boxes have 6, 9, 20 items each. Input is n, figure out if you can buy exactly n items.   It got me thinking that there is a number m, if n &gt;= m then you always can buy n items. If you generalize number of items in a box into s1 s2 s3 ... sy, number of boxes to buy into k1 k2 k3 ... ky, number of items you want to buy into n.  k1*s1 + k2*s2 + ... + ky*sy = n  How do I find the value m, if n &gt;= m there's always gonna be k1, k2, ..., ky which satisfies above equation. So that the algorithm can say yes you can without doing anything if n is large enough.
43369
During a period of $n$ days, two persons drink beers, one each. There are $n$ different beers $B_1, B_2, B_3,\ldots, B_n$, where $n\geqslant 1$ is an integer. t Person1 drinks the beers in order, i.e., on the $i$-th day, she drinks beer $B_i$. Person2 takes a uniformly random permutation $a_1, a_2, \ldots, a_n$ of the set $\{1, 2,\ldots, n\}$ and drinks beer $B_{a_i}$ on the $i$-th day. Let $X$ be the random variable whose value is the number of days during which person 1 and person 2 drink the same beer. Determine the expected value $\mathbb E(X)$ of $X$.  I know that I should let $X$ be the number of days during which both persons drink the same beer. And  $$ X_i = \left\{\begin{array}{rc} 1,&amp;\text{if they drink the same beer on the }i\text{-th day} \\ 0,&amp;\text{if they drink different beers on the }i\text{-th day}\end{array}\right. $$  Then  $$\begin{align} \mathbb E[X_i] &amp;= 1\cdot P(X=1) = P(\text{Person 1 and Person 2 drink beer }i) \\&amp;= P(\text{Person1 drinks beer }i)P(\text{Person 2 drinks beer }i) \\ &amp;= 1\cdot P(\text{Person 2 drinks beer }i) = \frac{1}{(n-i)!} \end{align}$$   So, is my approach right? If it's will the next step be: $\mathbb E(X) = n\cdot\frac{1}{(n-i)!}$?
481
It seems I put a   Bounty assigned by outside party, two lengthy, reference-citing answers, one "-1" (awarded the bounty), one "-2", two others scored "0" and "-2" respectively, the answers suggesting one or the other is correct, 73 comments and no consensus so far - and me, as the asker, lost without a clue what to think of the answers anymore -  this is no longer "Learners' Level" question, so I thought I'd bring it here and hear what the experts have to say.    Original question:  Plurality of verb depending on plurality of list elements  An edit was suggested to my sentence.     There was were an orange, some grapes, two apples and a small pile of cherries on the plate.   In my native language plurality of the verb always follows plurality of the first element on the list. There were an orange,... sounds awkward to me, no matter what follows up. My simple solution was reordering:     There were some grapes, an orange, two apples and a small pile of cherries on the plate.   But that's not the first time I faced this situation and I'd like to know what the rules of grammar say about that - was my editor overzealous or am I trying to copy rules of my language that don't apply in English?    Someone linked a  for negatives, where the situation is more clear-cut ("There is no..."). Same applies to connection with "or" apparently, per accepted answer there. Still, nothing for "and".  It seems there is a consensus that in that if the verb goes after the list, it will be plural.     an orange, some grapes, two apples and a small pile of cherries were on the plate.   There doesn't seem to be clear consensus for:     On the plate { was | were } an orange, some grapes, two apples and a small pile of cherries   Adding more to the confusion is the abbreviated form: "There's"     One last note. In North America, at least, there is a widespread use of using a singular form like "there is" and "there was", without regard for the subject item or items, and this "there is" is often shortened to "there's":           There's three apples on the table!      Could you please clarify this mess?
14682
There are 25 students, and every student gives a chocolate to 5 other students. Is it possible for every student to receive a chocolate from exactly the 5 students they've sent chocolate to?  My first though is to construct this as a graph, with each student as a vertex so there are 25 vertices. And each vertex has a minimum degree of 5 since every student is sending out 5 chocolate.   Am I on the right track? How should I proceed next?
115432
There are nine pairs of gloves each of different colors in how many ways can we select six gloves such that each is of different color? I tried like this : First number of ways in which we can select 6 colors out of nine it is $^9C_6$  ways. Then we can arrange it in 6! ways and also we can select either of two gloves out of the 6 in $2^6$ ways. Hence required number of ways should be $^9C_6$  .$2^6$.6!
119103
Omg this is driving me crazy seriously, it's a subproblem for a bigger problem, and i'm stuck on it.  Anyways i need the number of ways to pick $x[1]$ ammount of objects type $1$, $x[2]$ ammount of objects type $2$, $x[3]$ ammounts of objects type $3$ etc etc such that $$x[1] + x[2] + \ldots x[n] = k\;.$$ Order of the objects of course doesn't matter.  I know there is a simple formula for this something like $\binom{k + n}n$ or something like that i just can't find it anywhere on google.
71625
If I understand it correctly, nouns for fruits (and certain types of foodstuffs, such as pizza) are used as mass nouns if thought of as "food substance", rather than "portions". So is it correct that I say "I like apple" instead of "I like apples"? Is such reasoning as "I like apple as food, especially in cut pieces", just like "I like soup, pork, etc." (instead of "I like apples as items served as food, especially as whole fruits") valid, or at the very least, applicable in any ways?  I know there's , but the given answer is not satisfying to me. It flat out denies such a phrase as "I like apple" as grammatically incorrect. But is it though? If it is, when exactly do we use the mass nouns instead of the count nouns? Does the same principle go with "chicken" (the meat) and "chickens" the animal?  Edit: "Dog" is NOT a mass noun and the other question is completely irrelevant. This is absurd. My question concerns the use of mass nouns and count nouns for foods.
103206
How many &quot;$1$&quot; can be selected at most, provided that only one from each row and column of the attached matrix $A$ is selected?(That is, when one &quot;$1$&quot; is selected, no other element can be selected than the row and column where that &quot;$1$&quot; is located.)  Answer: I think $6$ can be chosen.  How can I give this answer in connection with graph theory? I think it has to do with the Latin Square(or Rectangle). But I can't connect and explain. Please help me. I thought a lot, but I could not establish the relationship completely. Thanks...
33046
what is the total number of distinct arrangements of $n$ oranges and $n$ apples around a round table? I have no idea how to go about.
75484
5 balls of different colours are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. The number of ways in which we can place the balls in the boxes so that no box remains empty.  My attempt:-  First choose 3 balls to be placed in 3 boxes so that none of them remain empty in ${{5}\choose{3}}\cdot3! = 60$ ways.  Now remaining 2 balls can go into any of the 3 boxes in $3\cdot3 = 9$ ways.  Total number of ways $= 60\cdot9 = 540$.  Where am I going wrong ?
60096
15 coupons are numbered 1,2,3...15. 7 coupons are selected at random one at a time with replacement. Find probability that 9 is the largest number appearing on the coupon. I am having a problem finding $N(a)$ for this problem.  My working:  $N(s)= 15^7$ (since you have 15 choices for each time you pick up a coupon)  $N(a)= 9^7$  (since you have 9 permissible choices for each time you pickup a  coupon)  But according to my textbook $N(a)= 9^7 -8^7$  but no further explanation is given.  So i would really appreciate it if someone could point out where i am going wrong. Thanks in advance :)