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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

Thank you so much sir but the  textbook says 520

$${Thank}\:{you}\:{so}\:{much}\:{sir}\:{but}\:{the} \\ $$$${textbook}\:{says}\:\mathrm{520} \\ $$

Commented by mrW1 last updated on 26/Dec/17

I′m quite sure that the answer is 560.

$${I}'{m}\:{quite}\:{sure}\:{that}\:{the}\:{answer}\:{is}\:\mathrm{560}. \\ $$

Commented by NECx last updated on 26/Dec/17

ok sir..... That′s the same thing I  got...  Thanks for your solution

$${ok}\:{sir}.....\:{That}'{s}\:{the}\:{same}\:{thing}\:{I} \\ $$$${got}... \\ $$$${Thanks}\:{for}\:{your}\:{solution} \\ $$$$ \\ $$

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