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xxyz

xyxz

xyzx

yxxz

yxzx

yzxx

if x is the digit that repeats then there are six possible placements of ten digits

y must be one of the remaining nine digits and z must be one of the further remaining eight digits

so 6x10x9x8 = 4320

(note: this supposes that leading zeros are included)

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Q: How many four digit numbers are there with two repeated digits?
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