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The number of functions \mathrm{f} rom the set \mathrm{A=\{0,1,2\}} in to the set \mathrm{B=\{0,1,2,3,4,5,6,7\}} such that \mathrm{f(i) \leq f(j)} for \mathrm{i<j\: and \: i, j \in A} is

Option: 1

{ }^{8} C_{3}


Option: 2

{ }^{8} C_{3}+2\left({ }^{8} C_{2}\right)


Option: 3

{ }^{10} C_{3}


Option: 4

{ }^{10} C_{4}


Answers (1)

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A function \mathrm{f: A \rightarrow B \: such\: that \: f\left ( 0 \right )\leq f\left ( 1 \right )\leq f\left ( 2 \right )} falls in one of the following four categories.

Case 1    \mathrm{ \quad f(0)<f(1)<f(2)}
There are \mathrm{{ }^{8} C_{3}} unctions in this category.

Case 2       \mathrm{\quad f(0)=f(1)<f(2)}
  There are \mathrm{{ }^{8} C_{2}}  functions in this category.

Case 3       \mathrm{\quad f(0)<f(1)=f(2)}
There are again \mathrm{{ }^{8} C_{2}} functions in this category.

Case 4      \mathrm{ \quad f(0)=f(1)=f(2)}
There are \mathrm{ { }^{8} C_{1}} functions in this category.

Thus, the number of desired functions is

\mathrm{{ }^{8} C_{3}+{ }^{8} C_{2}+{ }^{8} C_{2}+{ }^{8} C_{1}={ }^{9} C_{3}+{ }^{9} C_{2}={ }^{10} C_{3}}.

Posted by

Gautam harsolia

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