I have one 1p , one 2p , one 5p , one 10p and one 20p . How many different amounts can I make with them?

Answer 1

31 different amounts.

It is a question of combining the coins in different ways, knowing that no two possible combinations will give you the same amount.

The different combinations are :

Taking 1 coin at a time - 5 possible combinations
Taking 2 coins at a time - 10 possible combinations
Taking 3 coins at a time - 10 possible combinations
Taking 4 coins at a time - 5 possible combinations
Taking 5 coins at a time - 1 possible combination

This gives you 31 combinations / amounts in total.

You can check all possible combinations yourself to check.

Answer 2

1p + 2p = 3p
1p + 5p = 6p
1p + 10p = 11p
1p + 20p = 21p
2p + 5p = 7p
2p +10p = 12p
2p + 20p = 22p
5p + 10p = 15p
5p + 20p = 25p
10p + 20p = 30p
1p + 2p + 5p +10p + 20p = 38p
1p + 2p + 5p +10p = 18p
1p + 2p + 5p = 8p
2p + 5p +10p + 20p = 37p
5p +10p + 20p = 35p
1p + 5p +10p + 20p = 36p
1p + 2p +10p + 20p = 33p
1p + 2p + 5p + 20p = 28p
2p + 10p + 20p =32p
2p + 5p + 20p = 27p
2p + 5p + 10p = 17p
5p + 1p + 10p = 16p
5p + 1p + 20p = 26p
1p
2p
5p
10p
20p

And I give up I know there are more but I'm tired!

Answer 3

32 different amounts (if you include 0). if you don't allow 0 as an answer then 31. the answer is 2^5 or 2^5-1 depending on whether you allow 0 as a valid answer or not. the point is that you can choose to include each coin or not giving 2 choices per coin and hence 2^5 (2 to the power of 5) in total as there are 5 different coins. strictly speaking one then has to check that all the sums obtained are different but that is obvious looking at the denominations of the coins.

Answer 4

any 1 from 5 = 5
any 2 from 5 = 10
any 3 from 5 = 10
any 4 from 5 = 5
any 5 from 5 = 1

>>>>>>total = 31

but of course,you have the option of
not using any coins at all

therefore,you can make 32 different
amounts from the 5 coins including
nothing

{another way of looking at this
maximum is 38 minimum is 0
this gives 39 minus
(4,9,14,19,24,29 and 34)
=39-7=32}

i hope that this helps

Answer 5

12

Answer 6

It's easier than most are making it. Each coin has two possibilities, to be used or not. So the possiblities are 2*2*2*2*2 = 32 including the possiblility that none are used.

So, 31 amounts, not counting 0.

Answer 7

10 5 2 1
10 5 2
10 5 1
10 2 1
10 1
5 2 1
5 2
5 1
2 1
10
5
2
1

12 combos. Check to see if there are any that add up to the same amount.

Answer 8

15

Answer 9

It's the sum of the number of combinations of 5 amounts, taken 1 at a time, 2 at a time, 3 at a time, 4 at a time and 5 at a time.

Which is : 5C1 + 5C2 + 5C3 + 5C4 + 5C5

= 5! / (1! * 4!) + 5! / (2! * 3!) + 5! / (3! * 2!) + 5! / (4! * 1!) + 5! / (5! * 1!)

= 5 + 10 + 10 + 5 + 1

= 31

Answer 10

32