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Fundamental Principle of Counting

The number of...

Question

The number of odd proper divisors of $3^{p} . 6^{m} . 21^{n}$ is

A

(p + m + n + 1) (m + 1)(n + 1) 2

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B

(p + m + n + 1) (n + 1) – 1

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C

(p + m + n + 1) (n + 1) – 2

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D

(p + 1) (m + 1) (n + 1) – 2

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Solution

The correct option is B (p + m + n + 1) (n + 1) – 1
$3^{p} . 6^{m} . 21^{n} = 2^{m} . 3^{p + m + n} . 7^{n}$
$∴$ required number of proper odd divisors
= is the divisors of $3^{p + m + n} 7^{n}$ other than one = (p+m+n+1)(n+1) – 1

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Similar questions

\sum_{r = 1}^{n} r^{3} - \sum_{p = 1}^{n} \sum_{m = 1}^{p} \sum_{r = 1}^{m} 1 = 80

, then possible of

n

(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} +^{n} C_{n} x^{n}, x ϵ N

then prove that

^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - . . . . . . + (- 1)^{m - 1} \times^{n} C_{m - 1}

= (- 1)^{m - 1} \frac{(n - 1) (n - 2) . . (n - m + 1)}{(m - 1)!}

\sum_{m = 1}^{n} (\sum_{k = 1}^{m} (\sum_{p = k}^{m}^{n} C_{m} .^{m} C_{p} .^{p} C_{k})) =

n

is a positive integer, show that
(1)

n^{n + 1} - n {(n - 1)}^{n + 1} + \frac{n (n - 1)}{2!} {(n - 2)}^{n + 1} - \dots = \frac{1}{2} n (n + 1)!

n^{n} - (n + 1) {(n - 1)}^{n} + \frac{(n + 1) n}{2!} {(n - 2)}^{n} - \dots = 1

;
the series in each case being extended to

n

1^{n} - n 2^{n} + \frac{n (n - 1)}{1 \cdot 2} 3^{n} - \dots = {(- 1)}^{n} n!

{(n + p)}^{n} - n {(n + p - 1)}^{n} + \frac{n (n - 1)}{2!} {(n + p - 2)}^{n} - \dots = n!

;
the series in the last two cases being extended to

n + 1

n

p

+

ive integers such that

n \geq p + 2

D (n, p) ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^{n} C_{p} & ^{n} C_{p + 1} & ^{n} C_{p + 2}^{n + 1} C_{p} & ^{n + 1} C_{p + 1} & ^{n + 1} C_{p + 2}^{n + 2} C_{p} & ^{n + 3} C_{p + 1} & ^{n + 2} C_{p + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣

then prove that

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

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