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Question-114956

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Question Number 114956 by I want to learn more last updated on 22/Sep/20
Commented by I want to learn more last updated on 22/Sep/20
really?. But i cannot solve it sir.
$$\mathrm{really}?.\:\mathrm{But}\:\mathrm{i}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{sir}. \\ $$
Commented by PRITHWISH SEN 2 last updated on 22/Sep/20
Hmm really challenging one .
$$\mathrm{Hmm}\:\mathrm{really}\:\mathrm{challenging}\:\mathrm{one}\:. \\ $$
Commented by I want to learn more last updated on 22/Sep/20
Maybe mrW sir can see to it too
$$\mathrm{Maybe}\:\:\mathrm{mrW}\:\:\mathrm{sir}\:\mathrm{can}\:\mathrm{see}\:\mathrm{to}\:\mathrm{it}\:\mathrm{too} \\ $$
Commented by bobhans last updated on 22/Sep/20
=C_3 ^(43) ×C_3 ^(39) ×C_3 ^(35) ×C_3 ^(31) ×C_3 ^(27) ×C_3 ^(23) ×C_3 ^(19) ×C_3 ^(15) ×C_3 ^(11) ×C_3 ^7 ×C_3 ^3 = ((43.42.41)/(3.2.1))×((39.38.37)/(3.2.1))×((35.34.33)/(3.2.1))×((31.30.29)/(3.2.1))×((27.26.25)/(3.2.1))×((23.22.21)/(3.2.1))×((19.18.17)/(3.2.1))×((15.14.13)/(3.2.1))×((11.10.9)/(3.2.1))×((7.6.5)/(3.2.1))×1
$$={C}_{\mathrm{3}} ^{\mathrm{43}} ×{C}_{\mathrm{3}} ^{\mathrm{39}} ×{C}_{\mathrm{3}} ^{\mathrm{35}} ×{C}_{\mathrm{3}} ^{\mathrm{31}} ×{C}_{\mathrm{3}} ^{\mathrm{27}} ×{C}_{\mathrm{3}} ^{\mathrm{23}} ×{C}_{\mathrm{3}} ^{\mathrm{19}} ×{C}_{\mathrm{3}} ^{\mathrm{15}} ×{C}_{\mathrm{3}} ^{\mathrm{11}} ×{C}_{\mathrm{3}} ^{\mathrm{7}} ×{C}_{\mathrm{3}} ^{\mathrm{3}} \\ $$$$=\:\frac{\mathrm{43}.\mathrm{42}.\mathrm{41}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}×\frac{\mathrm{39}.\mathrm{38}.\mathrm{37}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}×\frac{\mathrm{35}.\mathrm{34}.\mathrm{33}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}×\frac{\mathrm{31}.\mathrm{30}.\mathrm{29}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}×\frac{\mathrm{27}.\mathrm{26}.\mathrm{25}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}×\frac{\mathrm{23}.\mathrm{22}.\mathrm{21}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}×\frac{\mathrm{19}.\mathrm{18}.\mathrm{17}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}×\frac{\mathrm{15}.\mathrm{14}.\mathrm{13}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}×\frac{\mathrm{11}.\mathrm{10}.\mathrm{9}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}×\frac{\mathrm{7}.\mathrm{6}.\mathrm{5}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}×\mathrm{1} \\ $$
Commented by I want to learn more last updated on 22/Sep/20
Thanks sir, but please can you explain the basis of your representation?
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{but}\:\mathrm{please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{the}\:\mathrm{basis}\:\mathrm{of}\:\mathrm{your}\:\mathrm{representation}? \\ $$

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