(A) Given the expression $\frac{n!}{r!(n-r)!}$ with $n=5$ and $r=2$. Substituting the values,we get $\frac{5!}{2!(5-2)!}$. This simplifies to $\frac{5

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(A) Given the expression $\frac{n!}{r!(n-r)!}$ with $n=5$ and $r=2$. Substituting the values,we get $\frac{5!}{2!(5-2)!}$. This simplifies to $\frac{5!}{2! \times 3!}$. Expanding the factorials,we have $\frac{5 \times 4 \times 3!}{2 \times 1 \times 3!}$. Canceling $3!$ from the numerator and denominator,we get $\frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10$.

Class 11 Mathematics Permutation and Combination Definition of combinations, Condition combinations

Easy

Evaluate $\frac{n!}{r!(n-r)!}$ when $n=5$ and $r=2$.

A
$10$
B
$20$
C
$30$
D
$40$

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Permutation and Combination

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Definition of combinations, Condition combinations

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If ${ }^{n-3} C_r + B \cdot { }^{n-3} C_{r-1} + B^{\prime} \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3} = { }^n C_r$ holds for all $n \geq r \geq 3$,then $(B, B^{\prime}) = $.

DifficultAP EAMCET 2022

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If $^{2n}C_2 : ^nC_2 = 9:2$ and $^nC_r = 10$,then $r = $

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Evaluate $\frac{n!}{r!(n-r)!}$ when $n=5$ and $r=2$.

Similar Questions

In how many ways can a team of $11$ players be formed out of $25$ players,if $6$ out of them are always to be included and $5$ are always to be excluded?

Out of $10$ white,$9$ black and $7$ red balls,the number of ways in which selection of one or more balls can be made,is

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If ${ }^{n-3} C_r + B \cdot { }^{n-3} C_{r-1} + B^{\prime} \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3} = { }^n C_r$ holds for all $n \geq r \geq 3$,then $(B, B^{\prime}) = $.

If $^{2n}C_2 : ^nC_2 = 9:2$ and $^nC_r = 10$,then $r = $

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