Question

A shop sells 6 different flavours of ice-creams. In how many ways can a customer choose 4 ice-cream cones if they contain only 2 or 3 different flavours.

Answer 1

Hint: Permutation means arrangement of any data in any sequence or order. In other words, the number of arrangements of n different things taken r at a time. It is denoted by \[ = \dfrac{{n!}}{{x!y!z!}}\]
\[{}^n{p_r} = n(n - 1)(n - 2)...(n - r + 1) = \dfrac{{n!}}{{(n - r)!}}\]
Combination is a way of selecting different items from a collection of items and in this the order of the items doesn’t matter. It is denoted by
\[{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]
In short Permutations means arrangement and combinations means selection.

Complete step-by-step answer:
Given,
Total flavours of ice cream available \[ = 6\]
And a customer chooses 4 ice-cream cones of 2 or 3 different flavours
Let us suppose flavours are \[\{ a,b,c,d,e,f\} \].

 Now if 6 ice creams of 3 different flavours are chosen
Then,

Let us suppose flavour ‘a’ occur 2 times
And ‘b’ & ‘c’ are other two flavours

∴ a,a,b,c are the chosen ice creams
Then no. of ways of selecting 4 ice creams of 3 different flavours is,

\[\begin{gathered}
  {}^6{C_3} \times \dfrac{{3!}}{{2!}} = \dfrac{{6!}}{{3!(3)!}} \times \dfrac{{3!}}{{2!}} \\
   \Rightarrow \dfrac{{6 \times 5 \times 4}}{2} = 60ways \\
\end{gathered} \]
Now if 6 ice creams of 2 different flavours are chosen
Then,

Let us suppose flavour ‘a’ occur 2 times and flavour ‘b’ occur 2 times

∴ a,a,b,b are the chosen ice creams
Then no. of ways of selecting 4 ice creams of 2 different flavours is,

\[\begin{gathered}
  {}^6{C_2} \times 3 = \dfrac{{6!}}{{4!(2)!}} \times 3 \\
   \Rightarrow \dfrac{{6 \times 5}}{2} \times 3 = 45ways \\
\end{gathered} \]

∴ Number of ways of selecting =2 different flavours+3 different flavours
                               \[ = 60 + 45 = 105\;ways.\]

Note: The no. of permutations of ‘n’ items taken all at a time, when x of them are similar and y of them are similar of another type and z are of similar of some other type and remaining are different
\[ = \dfrac{{n!}}{{x!y!z!}}\]