Question Number 218129 by mr W last updated on 30/Mar/25
$${how}\:{many}\:{different}\:{words}\:{can}\:{be} \\ $$$${formed}\:{from}\:{the}\:{word}\: \\ $$$$\boldsymbol{\mathrm{MATHEMATICS}}? \\ $$$${note}:\:\:{here}\:{a}\:{word}\:{should}\:{have}\:{at}\: \\ $$$${least}\:{two}\:{letters},\:{but}\:{mustn}'{t}\:{have}\:{a} \\ $$$${meaning}. \\ $$
Answered by vnm last updated on 30/Mar/25
![Σ_(n=2) ^8 C_8 ^n n!+3Σ_(n=0) ^7 C_7 ^n n!C_(n+2) ^2 + 3Σ_(n=0) ^6 C_6 ^n n!C_(n+4) ^2 C_(n+2) ^2 +Σ_(n=0) ^5 C_5 ^n n!C_(n+6) ^2 C_(n+4) ^2 C_(n+2) ^2 = =13938212 the same result: Σ_(n=2) ^(11) Σ_(k=0) ^(min(3, [(n/2)])) C_(8−k) ^(n−2k) (n−2k)!C_3 ^k Π_(i=0) ^(k−1) C_(n−2i) ^2](https://www.tinkutara.com/question/Q218145.png)
$$\underset{{n}=\mathrm{2}} {\overset{\mathrm{8}} {\sum}}{C}_{\mathrm{8}} ^{{n}} {n}!+\mathrm{3}\underset{{n}=\mathrm{0}} {\overset{\mathrm{7}} {\sum}}{C}_{\mathrm{7}} ^{{n}} {n}!{C}_{{n}+\mathrm{2}} ^{\mathrm{2}} + \\ $$$$\mathrm{3}\underset{{n}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}{C}_{\mathrm{6}} ^{{n}} {n}!{C}_{{n}+\mathrm{4}} ^{\mathrm{2}} {C}_{{n}+\mathrm{2}} ^{\mathrm{2}} +\underset{{n}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}{C}_{\mathrm{5}} ^{{n}} {n}!{C}_{{n}+\mathrm{6}} ^{\mathrm{2}} {C}_{{n}+\mathrm{4}} ^{\mathrm{2}} {C}_{{n}+\mathrm{2}} ^{\mathrm{2}} = \\ $$$$=\mathrm{13938212} \\ $$$${the}\:{same}\:{result}: \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\mathrm{11}} {\sum}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{min}\left(\mathrm{3},\:\left[\frac{{n}}{\mathrm{2}}\right]\right)} {\sum}}{C}_{\mathrm{8}−{k}} ^{{n}−\mathrm{2}{k}} \left({n}−\mathrm{2}{k}\right)!{C}_{\mathrm{3}} ^{{k}} \underset{{i}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}{C}_{{n}−\mathrm{2}{i}} ^{\mathrm{2}} \\ $$
Commented by mr W last updated on 31/Mar/25
$${great}!\:{thanks}\:{sir}! \\ $$
Answered by mr W last updated on 31/Mar/25
$${M},\:{A},\:{T}\:{can}\:{be}\:{used}\:{at}\:{most}\:{two}\: \\ $$$${times}\:{each}. \\ $$$${H},\:{E},\:{I},\:{C},\:{S}\:{can}\:{be}\:{used}\:{at}\:{most}\:{one} \\ $$$${time}\:{each}. \\ $$$$\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}\right)^{\mathrm{3}} \left(\mathrm{1}+{x}\right)^{\mathrm{5}} \\ $$$$=\mathrm{1}+\mathrm{8}{x}+\frac{\mathrm{59}{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{133}{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{\mathrm{409}{x}^{\mathrm{4}} }{\mathrm{4}}+\mathrm{113}{x}^{\mathrm{5}} \\ $$$$\:\:\:+\frac{\mathrm{735}{x}^{\mathrm{6}} }{\mathrm{6}}+\frac{\mathrm{441}{x}^{\mathrm{7}} }{\mathrm{8}}+\mathrm{24}{x}^{\mathrm{8}} +\frac{\mathrm{29}{x}^{\mathrm{9}} }{\mathrm{4}} \\ $$$$\:\:\:+\mathrm{11}{x}^{\mathrm{10}} +\frac{{x}^{\mathrm{11}} }{\mathrm{8}} \\ $$$${total}\:{number}\:{of}\:{different}\:{words}\: \\ $$$${with}\:\mathrm{2}\:{till}\:\mathrm{11}\:{letters}: \\ $$$$\frac{\mathrm{59}×\mathrm{2}!}{\mathrm{2}}+\frac{\mathrm{133}×\mathrm{3}!}{\mathrm{2}}+\frac{\mathrm{409}×\mathrm{4}!}{\mathrm{4}}+\mathrm{113}×\mathrm{5}! \\ $$$$\:\:\:+\frac{\mathrm{735}×\mathrm{6}!}{\mathrm{6}}+\frac{\mathrm{441}×\mathrm{7}!}{\mathrm{8}}+\mathrm{24}×\mathrm{8}!+\frac{\mathrm{29}×\mathrm{9}!}{\mathrm{4}} \\ $$$$\:\:\:+\mathrm{11}×\mathrm{10}!+\frac{\mathrm{11}!}{\mathrm{8}}=\mathrm{13}\:\mathrm{938}\:\mathrm{212}\:\checkmark \\ $$