Let $ P $ be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in $ P $ are formed by using the digits 1, 2 and 3 only, then the number of elements in the set $ P $ is :
Solution
<p>(i) number of numbers created using</p>
<p>$1111133=\frac{7!}{5!2!} \Rightarrow 21$</p>
<p>(ii) number of numbers created using</p>
<p>$1111223=\frac{7!}{4!2!} \Rightarrow 105$</p>
<p>(iii) number of numbers created using</p>
<p>$$\begin{aligned}
& 1112222=\frac{7!}{4!3!} \Rightarrow 35 \\
& \text { Total }=161
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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