The total number of ways in which $2 n$ persons can be divided into n couples is

\frac{2 n!}{n! n!}

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\frac{2 n!}{(2!)^{n}}

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\frac{2 n!}{n! (2!)^{n}}

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None of these

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Solution

The correct option is C $\frac{2 n!}{n! (2!)^{n}}$
Here, we are dividing $2 n$ people in $n$ groups of $2$ each and we are concerned with mere grouping. Hence, the required number of ways is $2 n$ people in $n$ groups of $2$ each, and we are concerned with mere grouping. Hence, the required number of ways is $\frac{2 n!}{n! (2!)^{n}}$ .

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Similar questions

Q. The number of ways of dividing

2 n

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n

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Q. The greatest coefficient in the expansion of

(x + \frac{1}{x})^{2 n}

The number of ways in which 2n different things can be divided equally into two distinct groups is $\frac{2 n!}{(n!)^{2}}$ (If order of the group is important)

Q. Prove that

1 - 2 n + \frac{2 n (2 n - 1)}{2!} - \frac{2 n (2 n - 1) (2 n - 1)}{3!} + . . . + (- 1)^{n - 1} \frac{2 n (n - 1) . . . (n + 2)}{(n - 1)}

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

where n is a + ive integer.

Q. The coefficient of

\frac{1}{x}

in the expansion of

{(1 + x)}^{n} {(1 + \frac{1}{x})}^{n}

is
(a)

\frac{n!}{[(n - 1)! (n + 1)!]}

(b)

\frac{(2 n)!}{[(n - 1)! (n + 1)!]}

(c)

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

(d) none of these

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The total number of ways in which 2n persons can be divided into n couples is

The correct option is C 2n!n!(2!)nHere, we are dividing 2n people in n groups of 2 each and we are concerned with mere grouping. Hence, the required number of ways is 2n people in n groups of 2 each, and we are concerned with mere grouping. Hence, the required number of ways is 2n!n!(2!)n.

The total number of ways in which $2 n$ persons can be divided into n couples is