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Standard XII

Mathematics

De Morgan's Law

A train consi...

Question

A train consists of n carriages and there are P passengers. Each one of the P passengers randomly selects the carriage in which he will ride. On the basis of above information, answer

$^{n} C_{1} 1^{p} -^{n} C_{2} 2^{p} +^{n} C_{3} 3^{p} . . . . + (- 1)^{n - 1 n} C_{n} n^{p}$ is equal to,

$(- 1)^{n - 1} n!, i f P < n$

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$0, i f P = n$

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$n!, i f P = n$

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$(- 1)^{n - 1} n!, i f P < n$

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Solution

The correct option is D
$(- 1)^{n - 1} n!, i f P < n$

$F (n, P) = \frac{n^{p} -^{n} C_{1} (n - 1)^{p} +^{n} C_{2} (n - 2)^{p} + . . . . (- 1)^{n - 1 n} C_{n - 1} 1^{p}}{n^{p}} F (4, 6) = \frac{4^{6} -^{4} C_{1} 3^{6} +^{4} C_{2} 2^{6} -^{4} C_{3} 1^{6}}{4^{6}} = \frac{195}{512}$
Also if P < n then probability is zero and if $P = n t h e n, \frac{n^{n} -^{n} C_{1} (n - 1)^{n} + . . . . + (- 1)^{n - 1 n} C_{n} 1^{n}}{n^{n}} = \frac{n!}{n^{n}}$

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Similar questions

A train consists of n carriages and there are P passengers. Each one of the P passengers randomly selects the carriage in which he will ride. On the basis of above information, answer

If n = 4 and p = 6, then the probability that there will be at least one passenger in each carriage is

Q. The number of odd proper divisors of

3^{p} . 6^{m} . 21^{n}

Q. If

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)!

;
(2)

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

terms; and
(3)

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

;
(4)

{(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

terms.

Q. If

n

and

p

be two

+

ive integers such that

n \geq p + 2

and

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)

Q. If

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

, then possible of

n

can be -

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A train consists of n carriages and there are P passengers. Each one of the P passengers randomly selects the carriage in which he will ride. On the basis of above information, answer nC11p−nC22p+nC33p....+(−1)n−1 nCnnp is equal to,

The correct option is D (−1)n−1n!,if P<n F(n,P)=np−nC1(n−1)p+nC2(n−2)p+....(−1)n−1 nCn−11pnpF(4,6)=46−4C136+4C226−4C31646=195512 Also if P < n then probability is zero and if P=n then,nn−nC1(n−1)n+....+(−1)n−1 nCn1nnn=n!nn

A train consists of n carriages and there are P passengers. Each one of the P passengers randomly selects the carriage in which he will ride. On the basis of above information, answer

$^{n} C_{1} 1^{p} -^{n} C_{2} 2^{p} +^{n} C_{3} 3^{p} . . . . + (- 1)^{n - 1 n} C_{n} n^{p}$ is equal to,