0
$\begingroup$

Suppose person X has $12$ dollars.In each of the first 5 days he buys one of the following items.

1.Item A for $1

2.Item B for $2

3.Item C for $3.

In how many ways can he spend the money during the first five days?

This is what I did.

Each of the five days he can buy Item A without running out of money. Also each of the five days he can buy Item B without running out of money.Therefore based on these 2 choices he has 2 selections each day.Thus $2^5$selections.
Now if he decides to buy item C,
If he buys Item C for 1 day then, still all other four days have 2 choices each day.Ways of selecting the day to buy item C=${5 \choose 1}$ .And each 4 days having 2 choices make the total number of selections=$2^4*5=80$.

If he buys Item C for 2 days then ways of selecting these two days =${5 \choose 1}$.All other three days have 2 choices each day.Total selections in this case=$2^3*10=80$.

If he buys Item C for 3 days then ways of selecting these two days =${5 \choose 2}$ There are 3 ways of selecting items for the next two days.Thus total number of selections for this case=$3*10=30$

Therefore total number of ways of spending 12 dollars is
$2^5+80+80+30=222$.

Is this correct.

$\endgroup$
  • $\begingroup$ Does order matter? Or does only the final shopping bag matter? $\endgroup$ – Alexandre Halm Oct 14 '14 at 11:19
  • 1
    $\begingroup$ It is correct if order matters. $\endgroup$ – Coolwater Oct 14 '14 at 11:58
1
$\begingroup$

So basically if order matters your universe is $\{1,2,3\}^5$ (cardinal $3^5=243$).

I would simply count how many of these quintuplets $(a_1,...,a_5) \in \{1,2,3\}^5$ do not satisfy $a_1+...+a_5 \le 12$ by distinguishing how many "$3$" values they have :

  • 5 times "3" : only one tuple $(3,3,3,3,3)$
  • 4 times "3" : fives tuples where "1" is the fifth value (from $(1,3,3,3,3)$ to $(3,3,3,3,1)$) and five ones where "2" is the last value
  • 3 times "3" : only those where the two other values are "2", I count $\binom{5}{2}$ ways to place the two "2"s

Answer : $243 - 1 - 5 - 5 - 10 = 222$.

$\endgroup$
0
$\begingroup$

If we understand the probleme as this:
What is the number of ways to spend 12 dollars in 5 days if we have to buy only one item a day, which costs 1 or 2 or 3 dollars?

It is the number of ways to partition 12 in 5 parts, with each of these 5 parts are 1 or 2 or 3:

$1 + 2 + 3 + 3 + 3 =12$
all the permutations: $\frac{5!}{3!} =20$

$2 + 2 + 2 + 3 + 3 =12$
all the permutations: $\frac{5!}{3!2!} =10$

So, $20+10 =30$ ways

Cheers

$\endgroup$
  • $\begingroup$ you don't need to spend the whole 12 dollars in the problem $\endgroup$ – Alexandre Halm Oct 14 '14 at 12:57
0
$\begingroup$

This is trickier, are number of sums to a limited quantity. You can calculate it using a generating function or by hand knowing the limitations.

Using a generating function will be the sum of the coefficients from the power 5 to 12 of this polynomial

$$f(x)=(x+x^2+x^3)^5=x^{15}+5x^{14}+15x^{13}+30x^{12}+45x^{11}+51x^{10}+45 x^9+30 x^8+15x^7+5 x^6+x^5\\ \to f^*(x)=30x^{12}+45x^{11}+51x^{10}+45 x^9+30 x^8+15x^7+5 x^6+x^5\\ f^*(1)=222$$

Im not consumist so I prefer the version where you can choose DONT buy if you want, in this version the generating function will be

$$g(x)=(1+x+x^2+x^3)^5=x^{15}+5 x^{14}+15 x^{13}+35x^{12}+65 x^{11}+101x^{10}+135 x^9+155 x^8+155x^7+135 x^6+101 x^5+65x^4+35 x^3+15 x^2+5 x+1\\ \to g^*(x)=35x^{12}+65 x^{11}+101x^{10}+135 x^9+155 x^8+155x^7+135 x^6+101 x^5+65x^4+35 x^3+15 x^2+5 x+1\\ g^*(1)=1003$$

;)

$\endgroup$
  • $\begingroup$ Does g represent the ordered selection and f the unordered $\endgroup$ – clarkson Oct 14 '14 at 12:51
  • $\begingroup$ @clarkson, all are ordered/unordered. Order doesnt matter for this problem. $g(x)$ is a version of the original question but considering that you can NOT buy too, i.e., spent 0$ if you want any day. $\endgroup$ – Masacroso Oct 14 '14 at 12:54
  • $\begingroup$ Everyday during these 5 days he buys something $\endgroup$ – clarkson Oct 14 '14 at 12:58
  • $\begingroup$ Guy, you read what I write? $\endgroup$ – Masacroso Oct 14 '14 at 12:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.