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Damek has five number cards lying on the table. There are two number cards with digit 1, two number cards with digit 2 and one number card with 0. How many different three-digit numbers can Damek form? (Number cannot being with 0).

(P.S. Is there a way to find the answer without listing all the possibilities?)

Sagot :

varshamittal029

Damek can form 14 three-digit numbers from the given situation. Hence, only 14 possibilities.

There are five number cards on the table.

2- digit 1 card

1- digit 0 card

2- digit 2 card

There is a possibility of putting 1 or 2 in the hundredth place.

If 1 is put in the hundredth place then there are 3 possibilities for tenth place 1,0,2

If 1 is put there then there is a possibility of 2 numbers 0,2 in ones place

If 2 is put then there is a possibility of 3 numbers 0,1,2 in ones place

If 0 is put then there is a possibility of 2 numbers 1,2 in ones place.

So, there are 7=(2+3+2) possibilities that the hundredth place is filled by 1.

Similarly, there will be 7 possibilities that the hundredth place is filled by 2.

Hence, there are 14 possibilities as required by the problem.

So, the possibilities of 3-digit numbers are (given the number cannot start with 0) 14.

Learn more about permutation here-

brainly.com/question/1216161

#SPJ10

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