for Bijou to get pleasure from

[sanju09]

There are five flavors of ice-cream: banana, chocolate, lemon, strawberry and vanilla, the ice-cream being in separate containers. Bijou can have three scoops of any possible grouping. How many choices will there be for Bijou to get pleasure from?
(A) 24
(B) 35
(C) 42
(D) 56
(E) 70



[spoiler]https://www.facebook.com/?ref=home[/spoiler]

[limestone]

My approach:
There're 5 flavors: A,B,C,D,E. A boy picked 3 times, and he can repick a flavor after it had been picked.
There're 3 cases that he can pick 3 scoops of ice-cream:
I. 3 different flavors
II. 2 same flavors, 1 other flavor
III.3 same flavors.

For case 1: As ABC is considered the same as CBA, use combination, total of choices 5C3 = 10

For case 2: use permutation, as the flavor being picked twice has different property with the flavor being picked once. In case 1, each flavor is picked once, they have the same property of being picked once, hence they can replace each other's place in the 3 picked scoops. While in case 2, AAB is different from BBA, although the same two flavors are picked. No. of Choices : 5P2 = 5*4 =20,
why 5P2, not 5P3, as I consider flavors to be picked, not scoops, here only 2 flavors are picked.

For case 3: 5P1 = 5 choices - AAA,BBB,CCC,DDD,EEE

Total choices: 10+20+5 = 35
Pick B.

[clock60]

here also got 35, but solved with formal approach.
if we have n- different types ( banana, lemon,.......-5 types.) and we need to made combination with k elements ( 3 scoops)
and the order of the things does not matter ( ban, lemon, vanila is the same as lemon banana vanila) and things can repeat ( vanila vanila,vanila, lemon lemon vanila.......) this can be done
(n+k-1)C(k)-ways
in our case (5+3-1)C(3)=7C3=35

[limestone]

if we have n- different types ( banana, lemon,.......-5 types.) and we need to made combination with k elements ( 3 scoops)
and the order of the things does not matter ( ban, lemon, vanila is the same as lemon banana vanila) and things can repeat ( vanila vanila,vanila, lemon lemon vanila.......) this can be done
(n+k-1)C(k)-ways
in our case (5+3-1)C(3)=7C3=35

Wow, I love formal approach, it's much shorter. But I will appreciate if you can prove how to come to this formulas or provide me with the sources of this formulas.

[sanju09]

if we have n- different types ( banana, lemon,.......-5 types.) and we need to made combination with k elements ( 3 scoops)
and the order of the things does not matter ( ban, lemon, vanila is the same as lemon banana vanila) and things can repeat ( vanila vanila,vanila, lemon lemon vanila.......) this can be done
(n+k-1)C(k)-ways
in our case (5+3-1)C(3)=7C3=35

Wow, I love formal approach, it's much shorter. But I will appreciate if you can prove how to come to this formulas or provide me with the sources of this formulas.

See that there are always 3 scoops of ice cream and 4 moves as you need to move 4 times to go from the first to the fifth container. By and large, there are r + (n - 1) spots and we want to choose r of them to have scoops, where n is the number of things to choose from, and you choose r of them with repetition allowed and order not mattering.

Formula comes this simple, choices are

= (n + r - 1)Cr

= (n + r - 1)!/[r! (n - 1)!]

In our question, n = 5 and r = 3, hence the many choices for Bijou to get pleasure from are

= (5 + 3 - 1)!/[3! (5 - 1)!]

=(7)!/[3! (4)!]

= [spoiler]35


B[/spoiler]

[clock60]

not sure that is can provide any proving of this formula, may be any link, but it is easy to find this material in net
combination with repetitions, or the number combinations with repetitions


https://en.wikipedia.org/wiki/Combination

https://www.csee.umbc.edu/~stephens/203/PDF/6-5.pdf
hope these links are valid