Question

A box contains $7$ red , $6$ white and $4$ blue balls . In how many ways can a selection of $3$ balls be made , if all the balls are red ?
A) $35$
B) $42$
C) $28$
D) $30$

Answer 1

We have to find the number of ways in which we can select $3$ balls and all of them are to be red. We solve this question using the concept of combinations of terms. We should also have the knowledge of the expansion of the factorials. We will get the total number of selections by choosing all the three balls from the given number of balls of red colour and hence applying the formula of combination, we will expand the expression and on simplifying, we get the value of the number of ways.

Complete step by step answer:
Given, A box contains $7$ red , $6$ white and $4$ blue balls .
Now, we have to select $3$ balls and all of them to be red.
Total number of ways of selection of the red balls, can be written as
Number of ways $= {}^7{C_3}$
Now we also know that the formula of combination can be written as
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Using the formula , we can write the expression for the number of ways as
Number of ways $= \dfrac{{7!}}{{3!\left( {4!} \right)}}$
On expanding the factorial , the expression can be written as
Number of ways $= \dfrac{{7 \times 6 \times 5 \times 4!}}{{3 \times 2 \times 1 \times \left( {4!} \right)}}$
Number of ways $= \dfrac{{7 \times 6 \times 5}}{6}$
On further simplifying , we get the value of number of ways as
Number of ways $= 35$
Hence, we can select all three red balls from the given balls in the bag in $35$ ways .
Therefore, option (A) is correct.

Note:
We can also directly calculate the value of the probability by making cases of boys and girls outcomes like the ones we make in the case of tossing a coin . Using this method we won’t have to use the concept of permutation and combinations .
Also , some formulas used :
${}^n{C_0} = 1$
${}^n{C_1} = n$
${}^n{C_2} = \dfrac{{n(n - 1)}}{2}$
${}^n{C_n} = 1$