The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is _______.
Answer (integer)
64
Solution
<p>$$\begin{array}{|c|c|c|}
\hline \mathrm{x} & \mathrm{y} & \mathrm{z} \\
\hline
\end{array}$$</p>
<p>Let $\mathrm{x}=2 \Rightarrow \mathrm{y}+\mathrm{z}=13$</p>
<p>$(4,9),(5,8),(6,7),(7,6),(8,5),(9,4), \rightarrow 6$</p>
<p>Let $x=3 \rightarrow y+z=12$</p>
<p>$(3,9),(4,8), \ldots \ldots . .,(9,3) \rightarrow 7$</p>
<p>Let $x=4 \rightarrow y+z=11$</p>
<p>$(2,9),(3,8), \ldots \ldots \ldots,(9,1) \rightarrow 9$</p>
<p>Let $x=5 \rightarrow y+z=10$</p>
<p>$(1,9),(2,8), \ldots \ldots . .,(9,1) \rightarrow 10$</p>
<p>Let $x=6 \rightarrow y+z=9$</p>
<p>$(0,9),(1,8), \ldots \ldots . .,(9,0) \rightarrow 9$</p>
<p>Let $\mathrm{x}=7 \rightarrow \mathrm{y}+\mathrm{z}=8$</p>
<p>$(0,9),(1,7), \ldots \ldots . .,(8,0) \rightarrow 9$</p>
<p>Let $x=8 \rightarrow y+z=7$</p>
<p>$(0,7),(1,6), \ldots \ldots . .,(7,0) \rightarrow 8$</p>
<p>Let $x=9 \rightarrow y+z=6$</p>
<p>$(0,6),(1,5), \ldots \ldots \ldots,(6,0) \rightarrow 7$</p>
<p>Total $=6=7+8+9+10+9+8+7=64$</p>
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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