In a lottery 5 different numbers are chosen from the first 90 positive integers. how many outcomes are there with the property that the last digits of all five numbers are different? (the last digit of 5 is 5 and the last digit of 34 is 4).

There are 1752574320 outcomes

Step-by-step explanation:

The option are:

1      11      21      31      41      51      61      71      81      91

2     12     .         .         .         .        .         .        .         .    

3     13     .         .         .         .        .         .        .         .

4     14     .         .         .         .        .         .        .         .

5     15     .         .         .         .        .         .        .         .

6     16     .         .         .         .        .         .        .         .

7     17     .         .          .         .        .         .        .         .

8     18     .         .          .         .        .        .         .         .

9     19     .         .          .         .        .        .         .         .

10    20    .         .          .         .        .        .         .         .

So we have to select 5 numbers, with  the property that the last digits of all five numbers are different

__ __ __ __ __

1. Have 90 options

⇒ 90

2. Have 90 - 9 options (e.g. if 2 was the first chosen number, then you can't select more 2, 12, 22, 32, 42, 52, 62, 72 or 82 )

⇒ 90 - 9 = 81

3. Have 90 - 9 that can't no be chosen more because share the same last number as the first number - 9 that can't no be chosen more because share the same last number as the second number

⇒ 90 - 9 - 9 = 72

4. Have 90 - 9 that can't no be chosen more because share the same last number as the first number - 9 that can't no be chosen more because share the same last number as the second number - 9 that can't no be chosen more because share the same last number as the third number

⇒ 90 - 9 - 9 - 9 = 63

5. Have 90 - 9 that can't no be chosen more because share the same last number as the first number - 9 that can't no be chosen more because share the same last number as the second number - 9 that can't no be chosen more because share the same last number as the third number - 9 that can't no be chosen more because share the same last number as the fourth number

⇒ 90 - 9 - 9 - 9 - 9 = 54

So now the number of possible combination with the given restriction is equal to the multiplication of the amount of option for the selection of each number (90 for the selection of the first, 81 for the selection of the second, 72 for the selection of the third number, 63 for the selection of the fourth and 54 for the selection of the fifth)

C= 90*81*72*63*54 = 1752574320

There are 1752574320 outcomes

question:

could we have options?

irrational, cannot