A-number-of-four-different-digits-is-formed-by-using-the-digits-1-2-3-4-5-6-7-in-all-possible-ways-each-digit-occuring-once-only-find-how-many-of-them-are-greater-than-3400-

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Question Number 26477 by NECx last updated on 26/Dec/17

A number of four different digits is formed by using the digits 1,2,3,4,5,6,7,in all possible ways each digit occuring once only.find how many of them are greater than 3400

$${A}\:{number}\:{of}\:{four}\:{different} \\ $$$${digits}\:{is}\:{formed}\:{by}\:{using}\:{the}\: \\ $$$${digits}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},{in}\:{all}\:{possible} \\ $$$${ways}\:{each}\:{digit}\:{occuring}\:{once} \\ $$$${only}.{find}\:{how}\:{many}\:{of}\:{them}\:{are} \\ $$$${greater}\:{than}\:\mathrm{3400} \\ $$

Answered by mrW1 last updated on 26/Dec/17

type 1: 3XYY with X=4,5,6,7 N_1 =4×5×4=80 type 2: ZYYY with Z=4,5,6,7 N_2 =4×6×5×4=480 ⇒N=80+480=560

$${type}\:\mathrm{1}:\:\mathrm{3}{XYY}\:{with}\:{X}=\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7} \\ $$$${N}_{\mathrm{1}} =\mathrm{4}×\mathrm{5}×\mathrm{4}=\mathrm{80} \\ $$$${type}\:\mathrm{2}:\:{ZYYY}\:{with}\:{Z}=\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7} \\ $$$${N}_{\mathrm{2}} =\mathrm{4}×\mathrm{6}×\mathrm{5}×\mathrm{4}=\mathrm{480} \\ $$$$\Rightarrow{N}=\mathrm{80}+\mathrm{480}=\mathrm{560} \\ $$

Commented by NECx last updated on 26/Dec/17