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Question: A shop sells 6 different flavours of ice-cream. In how many ways can a customer choose 4 ice-cream c...

A shop sells 6 different flavours of ice-cream. In how many ways can a customer choose 4 ice-cream cones if they are not necessarily of different flavours.

Explanation

Solution

We start solving the problem by recalling the fact that the total number of ways to choose r objects from the given n objects if the repetitions are allowed as n+r1Cr1{}^{n+r-1}{{C}_{r-1}}. We then compare the given problem with the statement and find the values of n and r. We then substitute this in the formula and use nCr=n!r!(nr)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} and n!=n×(n1)×(n2)×(n3)×......×2×1n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times ......\times 2\times 1. We then make necessary calculations required to get the required answer.

Complete step by step answer:
According to the problem, we have given a shop which sells 6 different flavours of ice-cream. We need to find the total number of ways that a customer can choose 4 ice-cream cones if they are not necessarily of different flavours.
We know that the total number of ways to choose r objects from the given n objects if the repetitions are allowed as n+r1Cr{}^{n+r-1}{{C}_{r}} (if 0 is not allowed).
Here we have to arrange 4 cones by choosing the flavours from the given six flavours. So, we get n=6n=6, r=4r=4.
So, the total number of ways the customer can choose 4 four ice-cream cones is 6+41C4=9C4{}^{6+4-1}{{C}_{4}}={}^{9}{{C}_{4}}.
We know that nCr=n!r!(nr)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} and n!=n×(n1)×(n2)×(n3)×......×2×1n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times ......\times 2\times 1.
So, we get 9C4=9!4!(94)!{}^{9}{{C}_{4}}=\dfrac{9!}{4!\left( 9-4 \right)!}.
9C4=9!4!5!\Rightarrow {}^{9}{{C}_{4}}=\dfrac{9!}{4!5!}.
9C4=(9×8×7×6×5×4×3×2×1)(4×3×2×1)×(5×4×3×2×1)\Rightarrow {}^{9}{{C}_{4}}=\dfrac{\left( 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \right)}{\left( 4\times 3\times 2\times 1 \right)\times \left( 5\times 4\times 3\times 2\times 1 \right)}.
9C4=9×8×7×64×3×2×1\Rightarrow {}^{9}{{C}_{4}}=\dfrac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}.
9C4=9×7×2\Rightarrow {}^{9}{{C}_{4}}=9\times 7\times 2.
9C4=126\Rightarrow {}^{9}{{C}_{4}}=126 ways.
So, we have that the customer can choose four ice-cream cones in 126 ways.

So, the correct answer is “126 Ways”.

Note: We should not make calculation mistakes while solving this problem. We can also solve this problem by checking the total number of ways to choose with only one flavour, two flavours, three flavours etc. To get the answer in this way we should do calculation perfectly with conscience. We should not consider choosing zero flavours as we are assuming that cones are sold only with the ice-cream.