Question

Given 6 different toys of red colour, 5 different toys of blue colour and 4 different toys of green colour. Combination of toys that can be chosen taking at least one red and one blue toy are:
(a) 31258
(b) 31248
(c) 31268
(d) None of these

Answer 1

Hint: First find the number of toys of each colour. Assign a variable to each. Now for each toy there are 2 possibilities it may be selected it may not be selected. By using this concept find the total number of combinations of selection. Now by using the product rule calculate total ways. So, there is one case from the number. Now again apply the rule of product to get the final answer.

Complete step-by-step answer:
Rule of sum: In combinations, the rule of sum or addition principle is basic counting principle. It is simply defined as, if there are A ways of doing P work and B ways of doing Q work. Given P, Q words cannot be done together. Total number of ways to do P, Q are given by (A+B) ways.
Rule of product: In combinations, the rule of product or multiplication principle is basic counting principle. It is simply defined as, if there are A ways of doing P work and B ways of doing Q work. Given P, Q works can be done at a time. Total number of ways to do both P, Q work are given by (A.B) ways.

Let us assume the variable gr for green toys in question. Number of toys in the gr event are given by:
gr= 4
So, each toy has two possibilities, selected or not selected. By product rule we must multiply 2, 4 times, we get
$n\left( gr \right)={{2}^{4}}...........(i)$
There is no condition of gr. So, n(gr) is fixed.
Let us assume the variable R for red toys in question. Number of toys in the R event are given by:
R=6
So, each toy has 2 possibilities, selected or not selected by product rule, we must multiple 2, 6 times, we get:
$R={{2}^{6}}$
We must select at least 1 red toy in our combination. So, we must subtract 1 red toy in our combination. So, we must subtract event of not selecting any red toys, we get:
$n(R)={{2}^{6}}-1.......(ii)$
Let us assume the variable B for blue toys in the question. Number of toys corresponding to the event B, are given by:
B=5
So, each toy has 2 possibilities, selected or not selected.
By product rule we must multiple 2, 5 times, we get:
$B={{2}^{5}}$
We must select at least one blue toy in our combination. So, subtract the event of not selecting any blue toy, we get:
$n\left( B \right)={{2}^{5}}-1.........(iii)$
By product rule, we say $\left( i \right)\times \left( ii \right)\times \left( iii \right)$ is required result, we get:
$N=n\left( gr \right)n\left( R \right)n\left( B \right)$
By substituting all the values, we get the equation as:
$N={{2}^{4}}\left( {{2}^{6}}-1 \right)\left( {{2}^{5}}-1 \right)$
By simplifying powers, we get value of N as:
$N=16\times 63\times 31$
By simplifying the product of last two terms, we get it as:
$N=16\times 1953$
By simplifying the above equation more, we get the value as:
N= 31248
So, there are 31248 ways possible satisfying a given condition.

Note: The idea of taking 2 possibilities bound situations and using product rules is very crucial. Generally, students forget to subtract 1 and get the wrong answer. And another misconception applied is -1 is done on the power like ${{2}^{a-1}}$ but it is wrong you must subtract only 1 event i.e. ${{2}^{a}}-1$. So, do carefully.