Question

The number of ways selecting $10$ balls out of an unlimited number of white, red, blue and green balls is
$A)256$
$B)286$
$C)84$
$D)$ None of these

Answer 1

Hint: First, let us see the permutation combination concept.
Since the question is to find the number of ways, so we are going to use permutation and combination methods which we will study on our schools to approach the given questions to find the number of ways since the number of permutations of r-objects can be found from among n-things is ${}^n{p_r}$(number of arrangements) where p refers to the permutation. Also, in Combination we have r-things and among n-things are ${}^n{c_r}$ which is the number of ways.
Formula used:
${}^{n + r - 1}{C_{r - 1}}$ is the formula to find the number of ways to pick the white, red, blue, and green balls among the ten balls

Complete step-by-step solution:
Since given that we have a number of ways to select $10$ balls out of an unlimited number of white, red, blue, and green balls, which means in each color the at least one ball needs to select because we are given that all four-color balls also, we can select at most unlimited different color balls.
Hence applying the combination formula for the number of ways with $10$ balls can be distributed to the four-color balls is ${}^{n + r - 1}{C_{r - 1}}$, where $n = 10$are the total count balls and $r = 4$ is the number of colors and each of them will receive at least one and thus we have, $r - 1 = 4 - 1 = 3$
Therefore, we get ${}^{n + r - 1}{C_{r - 1}} = {}^{10 + 3}{C_3} \Rightarrow {}^{13}{C_3}$
Now by applying the general combination formula, which is ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Then, we get ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}} \Rightarrow {}^{13}{C_3} = \dfrac{{13!}}{{3!(13 - 3)!}}$
Further solving we have, ${}^{13}{C_3} = \dfrac{{13!}}{{3!(10)!}} \Rightarrow \dfrac{{13 \times 12 \times 11 \times 10!}}{{3 \times 2 \times 10!}}$
Canceling the common terms, we have ${}^{13}{C_3} = \dfrac{{13 \times 12 \times 11}}{{3 \times 2}} \Rightarrow 286$
Thus, the number of ways to select $10$ balls out of an unlimited number of white, red, blue, and green balls is $286$
Therefore, the option $B)286$ is correct.

Note: Since the factorial can be expressed as $n! = n(n - 1)(n - 2).....2.1$ so make use of this and solved the combination method.
If the question is about the number of arrangements of the ten balls into the four sets of boxes or color balls then we may apply the permutation formula, which is ${}^n{p_r} = \dfrac{{n!}}{{(n - r)!}}$ for the permutation.