If a, b and c are greatest value of ^{19} mathrm{C}_{mathrm{p}},^{20} mathrm{C}

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6.Permutation and Combination

hard

If $a, b$ and $c$ are the greatest value of $^{19} \mathrm{C}_{\mathrm{p}},^{20} \mathrm{C}_{\mathrm{q}}$ and $^{21 }\mathrm{C}_{\mathrm{r}}$ respectively, then

$\frac{a}{11}=\frac{b}{22}=\frac{c}{21}$

$\frac{\mathrm{a}}{10}=\frac{\mathrm{b}}{11}=\frac{\mathrm{c}}{21}$

$\frac{\mathrm{a}}{10}=\frac{\mathrm{b}}{11}=\frac{\mathrm{c}}{42}$

$\frac{a}{11}=\frac{b}{22}=\frac{c}{42}$

(JEE MAIN-2020)

Solution

$a=^{19} \mathrm{C}_{10}, \mathrm{b}=^{20} \mathrm{C}_{10}$ and $\mathrm{c}= ^{21} \mathrm{C}_{10}$

$\Rightarrow \mathrm{a}=^{19} \mathrm{C}_{9}, \mathrm{b}=2\left(^{19} \mathrm{C}_{9}\right)$ and $\mathrm{c}=\frac{21}{11}\left(^{20} \mathrm{C}_{10}\right)$

$\Rightarrow \mathrm{b}=2 \mathrm{a}$ and $\mathrm{c}=\frac{21}{11} \mathrm{b}=\frac{42 \mathrm{a}}{11}$

$\Rightarrow a: b: c=a: 2 a: \frac{42 a}{11}=11: 22: 42$

Standard 11

Mathematics

The number of arrangements of the letters of the word $SATAYPAUL$ such that no two $A$ are together and middle letter is consonant, is

normal

If $^n{C_r} = {\,^n}{C_{r – 1}}$ and $^n{P_r}{ = ^n}{P_{r + 1}}$, then the value of $n$ is

medium

There are $m$ men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by $84,$ then the value of $m$ is

hard

(JEE MAIN-2019)

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medium

If $^n{C_r} = 84,{\;^n}{C_{r – 1}} = 36$ and $^n{C_{r + 1}} = 126$, then $n$ equals

medium

If $a, b$ and $c$ are the greatest value of $^{19} \mathrm{C}_{\mathrm{p}},^{20} \mathrm{C}_{\mathrm{q}}$ and $^{21 }\mathrm{C}_{\mathrm{r}}$ respectively, then

Solution

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