David's candy

Let the number of candies David has be $n$ . He can choose $2$ candies in $\binom{n}{2}$ ways. Thus we have the equation $\binom{n}{2}=210$ If you are familiar with combinations, you may recognize the answer from here; however, we can solve it algebraically using the expansion of $\binom{n}{2}$ , $\frac{n(n-1)}{2}$ . So the equation we had is really $\frac{n(n-1)}{2}=210\Rightarrow 0=n^2-n-420=(n-21)(n+20)$ David must have a positive number of candies, so we choose the positive root, $\boxed{n=21}$ , as our answer.

Let us say there are $x$ distinct candies.

Number of ways in which David can pick $2$ out of $x$ candies are $_{2}^{x}\textrm{C}$ .

Hence, $_{2}^{x}\textrm{C} = 210$

$\frac{x!}{2! \times (x - 2)!} = 210$

$\frac{x \times (x - 1) \times (x - 2)!}{2! \times (x - 2)!} = 210$

$\frac{x \times (x - 1)}{2!} = 210$

$x^{2} - x = 420$

$x^{2} - x - 420 = 0$

Solving this, we get

$x = 21$ and $x = -20$

Since $x$ has to be positive, $x = \boxed{21}$ is the answer!

Let the number of candies be $n$ . Then we have, \begin{align} {n \choose 2} &= 210 \\ \frac{n(n-1)}{2} &= 210 \\ n^2-n-420 &= 0 \\ (n-21)(n+20) &=0 \\ &\Rightarrow n = 21 \quad \text{or} \quad n = -20 \end{align} However you cannot have $-20$ candies so the final answer is $\fbox{21}$

n.C.2 = 210 ---> n! / (n-2)!.2! = 210 ---> n! / (n-2)! = 420 ----> n.(n-1).(n-2)! /(n-2)! = 420 ---> n.(n-1) = 420 ----> 420 = 21.20 = n(n-1). So, n = 21. Answer : 21

${x \choose 2}=210 \rightarrow \dfrac{x!}{2!(x-2)!}=210 \rightarrow \dfrac{x(x-1)(x-2)!}{2!(x-2)!}=210 \rightarrow \dfrac{x^{2}-x}{2}=210$

Solving $x^{2}-x-420=0$ we find $x=-20$ or $x=21$ but since $x \geq 0$ the answer is $\boxed{21}$

according to the question , he selected 2 candies out of n candies

then, nC2 = 210

on solving this we get n = 21

n
let total no . of candies be n. then no. of ways of selecting 2 candies= c =210
2 nc2=n!/(n-2)!/2! which is equal to n(n-1)/2 that implies, n square - n =210x2=420. we will get quadratic equation. on factorising we get n=21 only because in nc2, 'n' must be +ve integer

We can use the formula n(n-1)/2 = 210

Just take for example that David has 3 candies with different colors Red, Green and Blue. There will be 3 ways in which he can give his candies to his friend, namely, Red and Green, Green and Blue, and Red and Blue.

Calculated as: 3(3-1)/2 = 3

Thus in the problem above, n(n-1)/2 = 210 n(n-1) = 420 n^2 - n - 420 = 0

by getting the roots of 420, we can have a result of n = -20 and n = 21.

The final answer is 21.

because david has some distinct type of candies, Simply we can made to :

$\binom {x} {2}$ = $210$

$\frac {x!} {(x-2)! 2!}$ = $210$

$\frac {x!} {(x-2)!}$ = $420$ $\Rightarrow$ cause $420$ $=$ $21 \times 20$ , so $x$ $=$ $21$

And david has $\boxed {21}$ candies.

He wants to select 2 candies out of n and he can do it nC2 number of ways which is given as 210. nC2 = 210 n!/((n-2)!2!) = 210 n (n-1) = 420 n (n-1) = 21*20 By comparing we see that n = 21.

21 C2=210

suppose there are n candies the no. of ways to select any 2 of them will be nc2 which will be equal to 210 form quardatic equation the value of n comes out to be 21