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Question

The number of ways in which mn students can be distributed equally among m sections is-

(mn)ⁿ
$\frac{(m n)!}{(n!)^{m}}$
$\frac{(m n)!}{m!}$
$\frac{(m n)!}{m! n!}$

hint Hint:

In order to solve this question, we should have some knowledge regarding the concept of combination, that is for choosing r out of n items irrespective of their order we apply the formula, $C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial left parenthesis n minus r right parenthesis factorial end fraction$ . And therefore, we will choose n number of students for each section turn by turn.

The correct answer is: $fraction numerator left parenthesis m n right parenthesis factorial over denominator left parenthesis n factorial right parenthesis to the power of m end fraction$

Detailed Solution
In this question, we have been asked to find the number of ways in which we can distribute mn students equally among m section. Now, we have been given that there are mn students in total and they have to be distributed equally among m sections.
$S o comma space w e space c a n space s a y space i n space e a c h space s e c t i o n comma space t h e r e space w i l l space b e space fraction numerator m n over denominator m end fraction space equals space n space space s t u d e n t s. N o w comma space w e space t a k e space n space s t u d e n t s space f r o m space m n space s t u d e n t s space f o r space e a c h space s e c t i o n space b y space u sin g space t h e space f o r m u l a space o f space c o m b i n a t i o n comma space t h a t space i s comma scriptbase C subscript r space equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction end scriptbase presuperscript n$

$F o r space t h e space f i r s t space s e c t i o n comma space w e space c a n space c h o o s e space n space s t u d e n t s space o u t space o f space m n. space S o comma space w e space g e t space t h e space n u m b e r space o f space w a y s space o f space c h o o sin g space n space s t u d e n t s space f o r space t h e space f i r s t space s e c t i o n space a s space scriptbase C subscript n end scriptbase presuperscript m n end presuperscript space.$
For the second section, we will again choose n students but out of (mn – n) because n students have already been chosen for the first section. So, we get the number of ways of choosing n students for the second section as $scriptbase C subscript n end scriptbase presuperscript m n minus n end presuperscript$
Similarly, for the third section, we have to choose n students out of (mn – 2n). So, we get the number of ways of choosing n students for the third section as $scriptbase C subscript n end scriptbase presuperscript m n minus 2 n end presuperscript$
And we will continue it in the same manner up to all mn students will not be divided into m section.
So, for (m – 1)th section, we will choose n students from (mn – ( m – 2)n) student. So, we get the number of ways of choosing n students for (n – 1)th section, we get, $C presuperscript m n minus left parenthesis m minus 2 right parenthesis n end presuperscript subscript n$
And for the mth section, we get the number of ways for choosing students as, $C presuperscript m n minus left parenthesis m minus 1 right parenthesis n end presuperscript subscript n$

Hence, we can write the total number of ways of distributing mn students in m section as
$scriptbase C subscript n end scriptbase presuperscript m n end presuperscript cross times scriptbase C subscript n end scriptbase presuperscript m n minus n end presuperscript cross times scriptbase C subscript n end scriptbase presuperscript m n minus 2 n end presuperscript cross times.... cross times scriptbase C subscript n end scriptbase presuperscript m n minus left parenthesis m minus 2 right parenthesis n end presuperscript cross times scriptbase C subscript n end scriptbase presuperscript m n minus left parenthesis m minus 1 right parenthesis n end presuperscript$
$N o w comma space w e space w i l l space u s e space t h e space f o r m u l a space o f space scriptbase C r end scriptbase presuperscript n space space t o space e x p a n d space i t. space S o comma space w e space g e t comma$
$fraction numerator m n factorial over denominator n factorial left parenthesis m n minus n right parenthesis factorial end fraction cross times fraction numerator left parenthesis m n minus n right parenthesis factorial over denominator n factorial left parenthesis m n minus 2 space n right parenthesis factorial end fraction cross times space fraction numerator left parenthesis m n minus 2 n right parenthesis factorial over denominator n factorial left parenthesis m n minus 3 space n right parenthesis factorial end fraction cross times.... cross times fraction numerator left parenthesis m n minus left parenthesis m minus 2 right parenthesis n right parenthesis factorial over denominator n factorial left parenthesis m n minus left parenthesis m minus 1 right parenthesis n right parenthesis factorial end fraction cross times space fraction numerator left parenthesis m n minus left parenthesis m minus 1 right parenthesis n right parenthesis factorial over denominator n factorial left parenthesis m n minus m n right parenthesis factorial end fraction$

And we can further write it as,
$fraction numerator left parenthesis m n right parenthesis factorial over denominator n factorial end fraction cross times fraction numerator 1 over denominator n factorial end fraction cross times fraction numerator 1 over denominator n factorial end fraction cross times..... cross times 1 space equals space fraction numerator left parenthesis m n right parenthesis factorial over denominator left parenthesis n factorial right parenthesis to the power of m end fraction H e n c e comma space w e space c a n space s a y space t h a t space t h e space t o t a l space n u m b e r space o f space w a y s space o f space d i s t r i b u t i n g space m n space s t u d e n t s space i n space m space s e c t i o n space a r e space fraction numerator left parenthesis m n right parenthesis factorial over denominator left parenthesis n factorial right parenthesis m space end fraction.$

While solving this question, the possible mistake one can make is by always choosing n students for all sections from mn students which is totally wrong because at a time one student can only be in 1 section. So, if n students are selected for 1 section then in the second section, we will choose from (mn – n).

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The number of ways in which mn students can be distributed equally among m sections is-

(mn)n

In order to solve this question, we should have some knowledge regarding the concept of combination, that is for choosing r out of n items irrespective of their order we apply the formula, . And therefore, we will choose n number of students for each section turn by turn.

The correct answer is:

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(mn)ⁿ
$\frac{(m n)!}{(n!)^{m}}$
$\frac{(m n)!}{m!}$
$\frac{(m n)!}{m! n!}$

The correct answer is: $fraction numerator left parenthesis m n right parenthesis factorial over denominator left parenthesis n factorial right parenthesis to the power of m end fraction$

The line among the following that touches the $y to the power of 2 end exponent equals 4 a x$ is

The line among the following that touches the $y to the power of 2 end exponent equals 4 a x$ is

If the line $x plus y plus k equals 0$ is a tangent to the parabola $x to the power of 2 end exponent equals 4 y$ then k=

If the line $x plus y plus k equals 0$ is a tangent to the parabola $x to the power of 2 end exponent equals 4 y$ then k=