Question

The total number of six digit numbers $x _{1} x _{2} x _{3} x _{4} x _{5} x _{6}$ having having the property that $x _{1}< x _{2}< x _{3}< x _{4}< x _{5} \leq x _{6}$ is equal to

• A. ${ }^{10} C _{6}$
• B. ${ }^{12} C _{6}$
• C. ${ }^{11} C _{6}$
• D. None of these

Answer 1

Answer: (C)

We have to form six digit number $x_{1} x_{2} x_{3} x_{4} x_{5} x_{6}$ having $x _{1}< x _{2} \leq x _{3}< x _{4}< x _{5}< x _{6}$
So here arises following cases:
Case I : $x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6}$
Now single order is already fixed so we just neet to select 6 numbers out of 9 possible numbers $\{1,2,3,4$, $5,6,7,8,9\}$
zero is neglected because it can't be placed at any place
This can be done in ${ }^{9} C _{6}$ ways
Case II : $x_{1} < x_{2}=x_{3} < x_{4} < x_{5} < x_{6}$
Now we have to select 5 digit
No. of ways $={ }^{9} C _{5}$
Case III : $x _{1} < x _{2}= x _{3} < x _{4} < x _{5}= x _{6}$
Now we have to select 4 digits done in ${ }^{9} C _{4}$ ways.
Case IV: $x_{1} < x_{2} < x_{3} < x_{4} < x_{5}=x_{6}$
We have select 5 digits
which is done in $=\,{}^9 C_5$ ways.
Total no. ways $={ }^{9} C _{6}+2 \cdot{ }^{9} C _{5}+{ }^{9} C _{4}={ }^{9} C _{6}+{ }^{9} C _{5}+{ }^{9} C _{5}+$ ${ }^{9} C _{4}={ }^{10} C _{6}+{ }^{10} C _{5}={ }^{11} C _{6}$