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Question Number 198283 by cortano12 last updated on 16/Oct/23

$$\mathrm{Given}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{consisting}$$$$\mathrm{of}4\mathrm{digits}\mathrm{abcd}\mathrm{such}\mathrm{that}$$$$\mathrm{a}⩽\mathrm{b}⩽\mathrm{c}⩽\mathrm{d}\mathrm{is}...$$$$\left(\mathrm{A}\right)495\left(\mathrm{B}\right)385\left(\mathrm{C}\right)275$$$$\left(\mathrm{D}\right)165\left(\mathrm{E}\right)55$$

Commented by mr W last updated on 19/Oct/23

$$\left(A\right)495$$$$generallywithkdigits:\frac{(k+8)!}{8!k!}$$$$seeQ117061$$

Answered by MM42 last updated on 16/Oct/23

$$\{a,b,c,d\}\to \left(\begin{array}{c}9\\ 4\end{array}\right)=126$$$$\{a,a,b,c\}\to \left(\begin{array}{c}9\\ 3\end{array}\right)\times 3=252$$$$\{a,a,b,b\}\to \left(\begin{array}{c}9\\ 2\end{array}\right)=36$$$$\{a,a,a,b\}\to \left(\begin{array}{c}9\\ 2\end{array}\right)\times 2=72$$$$\{a,a,a,a\}\to 9$$$$$$$$ans=495✓\to A$$$$$$

Commented by cortano12 last updated on 17/Oct/23

$$\mathrm{why}(\mathrm{a},\mathrm{a},\mathrm{a},\mathrm{b})=\left(\begin{array}{c}9\\ 2\end{array}\right)\times 2?$$$$$$

Commented by mr W last updated on 17/Oct/23

$$(a,b,b,b)and(a,a,a,b)$$

Commented by MM42 last updated on 17/Oct/23

$$\underset{―}{\underbrace{⋚}}$$

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