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Question

Match the following :
A)nC0+nC1+nC2+....+nCn 1)(n+1)Pr
B)PrP(r−1) 2) 2n
C)nPr+rn.P(r−1) 3)(n+1)Pr
D)nPn4)n−r+1
5)n!



A
A2,B4,C1,D5
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B
A1,B4,C2,D5
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C
A2,B4,C5,D1
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D
A1,B3,C5,D2
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Solution

The correct option is A A2,B4,C1,D5
A: nC0+nC1+.....nCn=(1+1)n=2n

B: nPrnPr1=n!(nr+1)!n!(nr)!=nr+1

C: nPr+r.nPr1=n!(nr)!+r.n!(nr+1)!=n!(nr+1)!(nr+1+r)=(n+1)!(nr+1)!=n+1Pr

D: nPn=n!(nn)!=n!

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