Required favourable cases are (0,4),(0,8),(2,6),(2,10),(4,8),(6,10)
So, favourable cases =6
Total cases =11C2=211×10=55
Hence, the required probability =556
If two different numbers are taken from the set 0,1,2,3,.....,10; then the probability that their sum as well as absolute difference are both multiple of 4, is:
Held on 2 Apr 2017 · Verified 6 Jul 2026.
556
5512
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557
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