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2005 Paper 1 Q1
D: 1500.0 B: 1500.0
combinatorics digit sum stars and bars counting with constraints enumeration integer partitions

\(47231\) is a five-digit number whose digits sum to \(4+7+2+3+1 = 17\,\).

  1. Show that there are \(15\) five-digit numbers whose digits sum to \(43\). You should explain your reasoning clearly.
  2. How many five-digit numbers are there whose digits sum to \(39\)?

Solution:

  1. The largest a five-digit number can have for its digit sum is \(45 = 9+9+9+9+9\). To achieve \(43\) we can either have 4 9s and a 7 or 3 9s and 2 8s. The former can be achieved in \(5\) ways and the latter can be achieved in \(\binom{5}{2} = 10\) ways. (2 places to choose to put the 2 8s). In total this is \(15\) ways.
  2. To achieve \(39\) we can have: \begin{array}{c|l|c} \text{numbers} & \text{logic} & \text{count} \\ \hline 99993 & \binom{5}{1} & 5 \\ 99984 & 5 \cdot 4 & 20 \\ 99974 & 5 \cdot 4 & 20 \\ 99965 & 5 \cdot 4 & 20 \\ 99884 & \binom{5}{2} \binom{3}{2} & 30 \\ 99875 & \binom{5}{2} 3! & 60 \\ 99866 & \binom{5}{2} \binom{3}{2} & 30 \\ 98886 & 5 \cdot 4 & 20 \\ 98877 & \binom{5}{2} \binom{3}{2} & 30 \\ 88887 & \binom{5}{1} & 5 \\ \hline && 240 \end{array}

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1998 Paper 1 Q12
D: 1484.0 B: 1606.9
probability independent events combinatorics divisibility integer solutions stars and bars complementary counting

Suppose that a solution \((X,Y,Z)\) of the equation \[X+Y+Z=20,\] with \(X\), \(Y\) and \(Z\) non-negative integers, is chosen at random (each such solution being equally likely). Are \(X\) and \(Y\) independent? Justify your answer. Show that the probability that \(X\) is divisible by \(5\) is \(5/21\). What is the probability that \(XYZ\) is divisible by 5?

Solution: They are not independent: \begin{align*} && \mathbb{P}(X = 20 \,\, \cap Y = 20) = 0 \\ && \mathbb{P}(X = 20 )\mathbb{P}(Y = 20) \neq 0 \\ \end{align*} \begin{align*} X = 0: && 21 \text{ solutions} \\ X = 5: && 16 \text{ solutions} \\ X = 10: && 11 \text{ solutions} \\ X = 15: && 6 \text{ solutions} \\ X = 20: && 1 \text{ solutions} \\ 5 \mid X: && 55 \text{ solutions} \\ \\ && \binom{20+2}{2} = 11 \cdot 21 \text{ total solutions} \\ \Rightarrow && \mathbb{P}(5 \mid X) = \frac{55}{11 \cdot 21} = \frac{5}{21} \end{align*} \begin{align*} \mathbb{P}(5 \mid XYZ) &= 3\cdot \mathbb{P}(5 \mid X) - 2\mathbb{P}(5 \mid X, Y, Z) \\ &= \frac{3 \cdot 55 - 2 \cdot \binom{4+2}{2}}{11 \cdot 21} = \frac{35}{77} \end{align*}

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