5.b) Consider this in two steps:
Choose one jack from the four jacks.
Choose the other four cards in the hand.
Thus we have ( 41 )( 514 ) = 999600.
The error involves double counting. For instance you may have choosen the jack of hearts in the first selection and the jack of clubs as one of the cards in the second selection. With the above solution this hand would also be counted if you chose the jack of clubs in the first selection and the jack of hearts in the second selection.
To solve the problem correctly, it might be easiest to calculate the number of poker hands which contain no jacks and subtract this from the total number of 5-card poker hands.
| The number of hands containing no jacks is | ( |
48 | ). |
| 5 |
| The total number of hands which contain at least one jack is | ( |
52 | ) | - |
( |
48 | ) | = 886656. |
| 5 | 5 |