Let r and n be positive integers such that 1 - r - n- Then prove the following-a CrnCr -1 n - n - r -1 rb n -n-1 Cr-1-n-r-1n Cr-1c CrnCr-1n-1-nriv n Cr-2 -n Cr-1-n Cr-2-n-2 Cr-

(a) nbsp;nCrnCr 1 #160;= #160;n r+1r nbsp; LHS #160;= #160;nCrnCr 1 #160; #160; #160; #160; #160; #160; #160; #160; #160;= #16 ...

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Byju's Answer

Standard XII

Mathematics

Some Algebraic Implications of Inequalities

Let r and n b...

Question

Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:
(a) $\frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} = \frac{n - r + 1}{r}$
(b) n ·_{^{n −}}_¹C_r_{− 1} = (n − r + 1) ⁿC_r_{− 1}
(c) $\frac{{}^{n}C_{r}}{{}^{n - 1}C_{r - 1}} = \frac{n}{r}$
(iv) ⁿC_r + 2 · ⁿC_r_{− 1} + ⁿC_r_{− 2} = ^{n +}²C_r.

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Solution

(a)
$\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$

$LHS = \frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n!}{r! (n - r)!} \times \frac{(r - 1)! (n - r + 1)!}{n!} = \frac{(n - r + 1) (n - r)! (r - 1)!}{r (r - 1)! (n - r)!} = \frac{n - r + 1}{r} = RHS$

∴ $LHS = RHS$

(b)
$LHS = n .^{n - 1} C_{r - 1} = \frac{n (n - 1)!}{(r - 1)! (n - 1 - r + 1)!} = \frac{n!}{(r - 1)! (n - r)!} RHS = (n - r + 1)^{n} C_{r} = (n - r + 1) \frac{n!}{(r - 1)! (n - r + 1)!} = (n - r + 1) \frac{n!}{(r - 1)! (n - r + 1) (n - r)!} = \frac{n!}{(r - 1)! (n - r)!}$

∴ LHS = RHS

(c)
$\frac{n_{C_{r}}}{n - 1_{C_{r - 1}}} = \frac{n}{r}$

$LHS = \frac{^{n} C_{r}}{^{n - 1} C_{r - 1}} = \frac{n!}{r! (n - r)!} \times \frac{(r - 1)! (n - 1 - r + 1)!}{(n - 1)!} = \frac{n (n - 1)!}{r (r - 1)! (n - r)!} \times \frac{(r - 1)! (n - r)!}{(n - 1)!} = \frac{n}{r} = RHS$

∴ LHS = RHS

(d)
$^{n} C_{r} + 2 .^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$

$LHS =^{n} C_{r} + 2 .^{n} C_{r - 1} +^{n} C_{r - 2} = (^{n} C_{r} +^{n} C_{r - 1}) + (^{n} C_{r - 1} +^{n} C_{r - 2})$
$=^{n + 1} C_{r} +^{n + 1} C_{r - 1}$ [∵ $n_{C_{r}} + n_{C_{r - 1}} = n + 1_{C_{r}}$ ]
$=^{n + 2} C_{r}$ [∵ $n_{C_{r}} + n_{C_{r - 1}} = n + 1_{C_{r}}$ ]
= RHS
∴ LHS = RHS

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Similar questions

Let r and n be positive integers such that $1 \leq r \leq n .$ Then prove the following :

(i) $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
(ii) $n^{n - 1} C_{r - 1} = (n - r + 1)^{n} C_{r - 1}$
(iii) $\frac{^{n} C_{r}}{^{n - 1} C_{r - 1}} = \frac{n}{r}$
(iv) $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$

Q. Let n and r be two positive integers such that

n \geq r + 2

. Suppose

Δ (n, r) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^{n} C_{r} & ^{n} C_{r + 1} & ^{n} C_{r + 2}^{n + 1} C_{r} & ^{n + 1} C_{r + 1} & ^{n + 1} C_{r + 2}^{n + 2} C_{r} & ^{n + 2} C_{r + 1} & ^{n + 2} C_{r + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣

Show that

Δ (n, r) = \frac{^{n + 2} C_{3}}{^{n + 2} C_{3}} Δ (n - 2, r - 1)

Hence or otherwise,

Q. If

(2 \leq r \leq n),

then

^{n} C_{r} + 2 \cdot^{n} C_{r + 1} +^{n} C_{r + 2}

is equal to

Q. If

n

and

r

are positive integers such that

r < n

, then

^{n} C_{r} +^{n} C_{r - 1} =

Q. If n and r are two positive integers such that

n \geq r

, then

^{n} C_{r + 1} +^{n} C_{r} =

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Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following: (a) CrnCr-1n=n-r+1r (b) n · n − 1Cr − 1 = (n − r + 1) nCr − 1 (c) CrnCr-1n-1=nr (iv) nCr + 2 · nCr − 1 + nCr − 2 = n + 2Cr.