Match the following - A n C 0 - n C 1 - n C 2 - n C n 1 n-1 P r B P r P r-1 2 2n C nPr - r n -Pr-1 3 n-1Pr D nPn4 n-r-1 5 n-

The correct option is A A 2, B 4,C 1,D 5 A: ^nC_0+^nC_1+.....^nC_n=(1+1)^n=2^n B: ^nP_r^nP_r 1=n!(n r+1)!n!(n r)!=n r+1 C: ^nP_r+r. #160; ...

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Byju's Answer

Standard XII

Mathematics

Combination

Match the fol...

Question

Match the following :
$A)^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + . . . . +^{n} C_{n}$ 1) $^{(n + 1)} P_{r}$
$B) \frac{P_{r}}{P_{(r - 1)}}$ 2) $2^{n}$
$C)^{n} P_{r} + r^{n} . P_{(r - 1)}$ 3) $^{(n + 1)} P_{r}$
$D)^{n} P_{n}$ 4) $n - r + 1$
5) $n!$

A - 2, B - 4, C - 1, D - 5

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A - 1, B - 4, C - 2, D - 5

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A - 2, B - 4, C - 5, D - 1

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A - 1, B - 3, C - 5, D - 2

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Solution

The correct option is A $A - 2, B - 4, C - 1, D - 5$
A: $^{n} C_{0} +^{n} C_{1} + . . . . .^{n} C_{n} = (1 + 1)^{n} = 2^{n}$

B: $\frac{^{n} P_{r}}{^{n} P_{r - 1}} = \frac{n! (n - r + 1)!}{n! (n - r)!} = n - r + 1$

C: $^{n} P_{r} + r .^{n} P_{r - 1} = \frac{n!}{(n - r)!} + r . \frac{n!}{(n - r + 1)!} = \frac{n!}{(n - r + 1)!} (n - r + 1 + r) = \frac{(n + 1)!}{(n - r + 1)!} =^{n + 1} P_{r}$

D: $^{n} P_{n} = \frac{n!}{(n - n)!} = n!$

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$A)^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + . . . . +^{n} C_{n}$	1) $^{(n + 1)} P_{r}$
$B) \frac{P_{r}}{P_{(r - 1)}}$	2) $2^{n}$
$C)^{n} P_{r} + r^{n} . P_{(r - 1)}$	3) $^{(n + 1)} P_{r}$
$D)^{n} P_{n}$	4) $n - r + 1$
	5) $n!$

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!

The correct option is A A−2,B−4,C−1,D−5A: nC0+nC1+.....nCn=(1+1)n=2nB: nPrnPr−1=n!(n−r+1)!n!(n−r)!=n−r+1C: nPr+r.nPr−1=n!(n−r)!+r.n!(n−r+1)!=n!(n−r+1)!(n−r+1+r)=(n+1)!(n−r+1)!=n+1PrD: nPn=n!(n−n)!=n!

Match the following :
$A)^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + . . . . +^{n} C_{n}$ 1) $^{(n + 1)} P_{r}$
$B) \frac{P_{r}}{P_{(r - 1)}}$ 2) $2^{n}$
$C)^{n} P_{r} + r^{n} . P_{(r - 1)}$ 3) $^{(n + 1)} P_{r}$
$D)^{n} P_{n}$ 4) $n - r + 1$
5) $n!$