Question

If ${}^{\prime }abc{d}^{\prime }$ is a $4$ digit number where $a>b\ge c>d$, then the number of such numbers are

• A. ${}^9{C}_4+{}^9{C}_3$
• B. ${}^{11}{C}_4$
• C. ${}^{10}{C}_4$
• D. ${}^{10}{C}_4+{}^8{C}_3$

Answer 1

The correct option is B ${}^{11}{C}_4$

Given condition : $a>b\ge c>d\Rightarrow a>b>c>d$ or $a>b=c>d$

Case $1:a>b>c>d$
$\Rightarrow$ Required ways = Total combinations of $4$ digits selected from $0-9.$
(example : $0,1,2,3$ is one of the combination of required number as $3210$)
$\therefore$ Required ways $={}^{10}{C}_4$

Case $2:a>b=c>d$
This means we have to make combination of $3$ digits (say $1,2,3$) out of which one is repeated twice and that will give one combination as $\left(3221\right)$
$\therefore$ Required ways $={}^{10}{C}_3\times 1={}^{10}{C}_3$

Total required ways $={}^{10}{C}_4+{}^{10}{C}_3={}^{11}{C}_4$