Present Wrapping

There are $7$ children, and you've mistakenly bought $2 \times7 = 14$ toy trains. Now, you put one toy train into each child's gift basket, that is you have put $7$ toy trains and you are thus left with $7$ toy trains. Now that your lazy mode is on, you have to randomly put the remaining $7$ toy trains into $7$ gift baskets. It is the same as dividing $7$ stars by $7$ bars. So, by stars and bars method, we can get the total number of ways in which this can be done $= {7+7-1 \choose7-1} = {13 \choose6} = 1716 = N$ . Since the questions ask for the last three digits of $N$ , our answer is $\fbox{716}$ .

The problem is to distribute $14$ toy trains among $7$ children Bob, Bill, Barrie, Bart, Brandon, Bernadette and Bertha. The value of $N$ is the number of solution of the equation $b_1+b_2+b_3+b_4+b_5+b_6+b_7=14$ where $b_i \ge 1$ for $i=1, \ldots, 7$ .

So, $N=^{(14-1)}C_{(7-1)}=1716$ .

Answer is $\boxed{716}$

There are seven children and we bought twice the amount of (identical) trains necessary, so we got $14$ trains. In the first round, we give everyone $1$ train each, and since they are identical, it doesn't make a difference which train we gave whom. Now we have $7$ trains. We randomly distribute this to the children.

Hmm, so we have $7$ children, and $7$ trains. We distribute the $7$ trains randomly. Sounds a lot like the famous stars and bars problem. Let the seven trains be the stars and let the bars represent which child gets what. So, one possible configuration is

 *|*||**|**|*||

The figure above excludes only the extra seven trains. Initially, every child got $1$ identical train. So, in the figure above,

• First child: $1$ train
• Second child: $1$ train
• Third child: $0$ trains
• Fourth child: $2$ trains
• Fifth child: $2$ trains
• Sixth child: $1$ train
• Seventh child: $0$ trains

Other configurations like

****|*|*|*||||

are also possible. The number of possible configurations is given by ${n+k-1 \choose k}$ where $n$ is the number of bars and $k$ are the number of stars.

Now, in our example, $n = 7$ and $k = 7$ . Thus, our answer is ${7 + 7 - 1 \choose 7} = 1716$ and the last three digits of this number are $\boxed{716}$ .

Pretend like we have each train in a row. So we have a row of 14 trains. There are 13 total spaces between the trains.

Say we also have 6 random sticks that we took from the Christmas tree. If we place all 6 sticks in the 13 spaces, we've immediately divided all 14 trains among the 7 children, and each child gets at least one train. (What each child gets is equal to the number of trains before or after each stick, or between the set of sticks.)

Therefore, $N = {13 \choose 6} = 1716$ So the answer is $716$

Direct formula to put N things in R bags is C (N+R-1,R-1).

In this case, N=7, R=7, So, 13C6 = 1716.

Last 3 digits = 716

after putting 7 ,7 left so equavalent to no of ways of attaining sum=7,no of 13c7=1716 so reqd ans=176

remaining toy = 7
ways to put remaining trains toy in the baskets is 13C6 = 1716
so the last three digits of N is 716