Collection of Maths Problems

Word problems

Task number: 3865

• Variant

  The pastry shop sells three types of desserts – pinwheels, canapes and punch cakes. In how many ways can you buy 12 desserts so that at least two desserts are bought of each type but at most three canapes?

• Solution

  For pinwheels and punch cakes we take the polynomial \(x^2+x^3+x^4+…\). The fact that we start from the square corresponds to the fact that these desserts must be picked each at least twice.

  Aside note: We could quit these two series at \(x^8 \), because with a total of 12 desserts we would not be able to choose more than 8 pinwheels or punch cakes. If we leave the series infinite, the resulting polynomial will also work for total numbers of desserts other than 12.

  For the canapes we take the polynomial \(x^2+x^3\).

  We get the polynomial \((x^2+x^3+x^4+…)^2(x^2+x^3)=x^6\frac{1+x}{(1-x)^2}\).

  Now either determine the searched sequence directly \( (0{,}0,0{,}0,0{,}0,1{,}3,5{,}7,9,…) \), or by using of the generalized binomial theorem we get \(\displaystyle x^6\frac{1+x}{(1-x)^2}= \sum_{i=0}^\infty (2i+1)x^{i+6}\)

  We see that \(x^k\) is assigned the coefficient \(2(k-6)+1\).

• Answer

  The value sought is the coefficient by \(x^{12}\) in the product \((x^2+x^3+x^4+…)^2(x^2+x^3)\) and is equal to 13.

• Variant

  How many ways can 10 identical balloons be distributed between two boys and two girls so that each boy receives at least one balloon and each girl at least two?

• Hint

  Use the generalized binomial theorem to specify the coefficients.

• Solution

  For boys we take the polynomial \((x+x^2+x^3+…)\), for girls \((x^2+x^3+x^4+…)\).

  We get \(\displaystyle (x+x^2+x^3+…)^2(x^2+x^3+x^4+…)^2 =x^6\frac{1}{(1-x)^4} =\sum_{i=0}^\infty \binom{i+3}{3}x^{i+6}\).

• Answer

  The value sought is the coefficient by \(x^{10}\) in the product \((x+x^2+x^3+…)^2(x^2+x^3+x^4+…)^2\) and is equal to \(\binom{7}{3}=35\).

• Variant

  The beekeeper asks the orchardist to plant 25 new trees, which he has only 4 species available. Orchardist’s wife refuses a walnut tree because it is large and occupies a lot of space. (In addition, everybody knows that walnut honey can’t be eaten :)) Apple trees are also against her taste, they already have too many. On the other hand, she loves infinitely cherry-plum jam, so she imposes harsh conditions – at most one walnut tree, not more than 10 apple trees, at least 6 cherry trees and at least 8 plum trees (slivovitz – an intensive motivation) or divorce. In how many ways can the orchardist salvage his matrimony? (I.e. in how many different ways he can choose the tree species?)

• Solution

  The generating function is

  \( \displaystyle (1+x)(1+x+x^2+…+… x^{10})(x^6+x^7+…)(x^8+x^9+…)= \frac{x^{14}(1+x)(1-x^{11})}{(1-x)^3}= \left(x^{14}+x^{15}-x^{25}-x^{26}\right)\sum_{i=0}^\infty \binom{i+2}{i}x^i \)

  The coefficient by \(x^{25}\) is \(\binom{11+2}{2}+\binom{10+2}{2}-\binom{0+2}{2}=78+66-1=143\).

• Answer

  The orchardist can select the trees in 143 different ways.