Permutations and Combinations Test 44

$$\displaystyle \sum _{ 0\le i\le  }^{  }{ \sum _{ j\le 10 }^{  }{ ^{ 10 }{ { C }_{ j } } }  } \quad \quad ^{ j }{ { C }_{ i } }$$

Add upper suffixces if their sum is even the half of it will be the value r otherwise the value of r be sum of upper suffixces +1 divided by 2.
Given $$\displaystyle S=\sum_{t=0}^{m}\binom{10}{i}\binom{20}{m-i}=\binom{10}{0}\binom{20}{m}+\binom{10}{1}\binom{20}{m-1}+\binom{10}{2}\binom{20}{m-2}+...+\binom{10}{m}\binom{20}{0}$$
In such cases we observe the upper suffixces and lower suffixces.
Upper suffices are 10 and 20 and lower suffixces taken the value (0...m).
In such cases we consider the two Binomial expansions $$\displaystyle \left ( 1+x \right )^{p}$$ and $$\displaystyle \left ( 1+x \right )^{q}$$ (Here p=10, q=20 or q=10, p=20) multiply them and collect the required exponent of x to which the given expression can be obtained.
Consider $$\displaystyle \left ( 1+x \right )^{10}\left ( 1+x \right )^{20}=\left ( ^{10}C_{0}+^{10}C_{2}x^{2}+...+^{10}C_{10^{x^{10}}} \right )\left ( ^{20}C_{0} +^{20}C_{1}x+...+^{20}C_{20^{x^{20}}}\right )$$
$$\displaystyle \left ( 1+x \right )^{30}=\left ( ^{10}C_{0} +^{10}C_{1^{x^{1}}}+...+^{10}C_{10^{x^{10}}}\right ).\left ( ^{20}C_{0}+^{20}C_{1}x^{1}+...+^{20}C_{m^{x^{m}}}+...+^{20}C_{20^{x^{20}}} \right )$$
and collecting coefficient of $$\displaystyle x^{m}$$ both side, we get  $$\displaystyle ^{30}C_{m}=\binom{10}{0}\binom{20}{m}+\binom{10}{1}\binom{20}{m-2}+...+\binom{10}{m}\binom{20}{0}$$ now for max value of the expression will be at $$\displaystyle m=\frac{30}{2}=15$$ as upper suffix is even $$\displaystyle \therefore m=15$$ which is given in (a).

$$\displaystyle 2^{k}\binom{n}{0}\binom{n}{k}-2^{k-1}\binom{n}{1}\binom{n-1}{k-1}+2^{k-2}\binom{n}{2}\binom{n-2}{k-2}..+(-1)^{k}\binom{n}{k}\binom{n-k}{0}$$
We know that

A properties of combinations is
$$ ^nC_r = ^nC_{r-1} $$

So, $$^{100}C_{97}  = ^{100}C_{ 100-97 } =  ^{100}C_{ 3 } = 1,61,700 $$

Paths are shown as :

Similarly if we start from A towards B we get another $$4$$ paths.

Similarly if we start from A towards B again $$3$$ paths.

∴ Total different paths $$4\times3=12$$

$$ ^nC_r = \dfrac { n! }{ (r!)(n-r)! } $$

Given, $$ ^nC_3 =^nC_5 $$
$$ => \dfrac { n! }{ (3!) (n-3)! } =\dfrac { n! }{ 5! (n -5)! } $$
$$ => 3 ! \times (n - 3) \times (n - 4) \times (n-5) ! = 5 \times 4 \times 3 ! \times (n-5)! $$
$$ => (n - 3) \times (n - 4) = 5 \times 4 $$

From this, $$ n - 3 = 5 $$
$$ => n = 8 $$
 

We know that, $$ { { n }_{ C } }_{ r }=\dfrac { n! }{ r!(n-r)! } $$

Given, $$ \dfrac { { { 19 }_{ C } }_{ r } }{ { { 19 }_{ C } }_{ r-1 } } =\quad \dfrac { 2 }{ 3 } $$

$$ =>\dfrac { \dfrac { 19! }{ r!(19-r)! }  }{ \dfrac { 19! }{ (r-1)!(19-r+1)! }  } =\quad \dfrac { 2 }{ 3 } $$

$$ => \dfrac { (r-1)!(20-r)! }{ r!(19-r)! } =\quad \dfrac { 2 }{ 3 } $$

$$ =\dfrac { (r-1)!\times (20-r) \times (19-r)! }{ r\quad \times (r-1)!\times (19-r)! } =\quad \dfrac { 2 }{ 3 } $$
$$ => \dfrac { 20-r }{ r\quad  } =\quad \dfrac { 2 }{ 3 } $$
$$ => r = 12 $$

So,
$$ { { 14 }_{ C } }_{ 12 }=\dfrac { 14! }{ 12!(14-12)! } =\dfrac { 14! }{ 12!(2)! } =\dfrac { 14\times 13\times 12! }{ 12!\times 2 } =\quad 7\times 13=91 $$

According to the given condition:

The first person can choose to sit in any of the $$ 5 $$ seats in $$ 5 $$ ways.

Then the second person can choose to sit in any of the remaining $$ 4 $$ seats in $$ 4 $$ ways.

Lastly, the third person can choose to sit in any of the remaining $$ 3 $$ seats in $$ 3 $$ ways.

So, total number of ways $$ = 5 \times 4 \times 3 = 60 $$ ways