in a certain group of 10 developers, 4 developers code only in python and the rest program in either ruby on rails or php - but not both. if a developer organization is to choose a 3 - member team, which must have at least 1 developer who codes in python, how many different programming teams can be chosen?	Select the correct option:
A) 100  B) 40  C) 66  D) 80  E) 75
<gadget id="calculator">factorial(10)</gadget>
<output>3_628_800</output>

<gadget id="calculator">10 - 3</gadget>
<output>7</output>

<gadget id="calculator">factorial(7)</gadget>
<output>5_040</output>

<gadget id="calculator">factorial(3)</gadget>
<output>6</output>

<gadget id="calculator">5_040 * 6</gadget>
<output>30_240</output>

<gadget id="calculator">3_628_800 / 30_240</gadget>
<output>120</output>

<gadget id="calculator">10 - 4</gadget>
<output>6</output>

<gadget id="calculator">factorial(6)</gadget>
<output>720</output>

<gadget id="calculator">6 * 6</gadget>
<output>36</output>

<gadget id="calculator">720 / 36</gadget>
<output>20</output>

<gadget id="calculator">120 - 20</gadget>
<output>100</output>

<result>A</result>
in a certain group of 10 developers, 4 developers code only in python and the rest program in either ruby on rails or php - but not both. if a developer organization is to choose a 3 - member team, which must have at least 1 developer who codes in python, how many different programming teams can be chosen?
subtract(divide(factorial(10), multiply(factorial(subtract(10, 3)), factorial(3))), divide(factorial(subtract(10, 4)), multiply(factorial(3), factorial(3))))
two ways . . . 1 ) total ways = 10 c 3 = 10 ! / 7 ! 3 ! = 120 . . ways without python developer = 6 c 3 = 6 ! / 3 ! 3 ! = 20 . . ways of at least one python developer = 120 - 20 = 100 . . 2 ) ways of selecting only one = 4 * 6 c 2 = 4 * 15 = 60 . . ways of selecting only two = 4 c 2 * 6 c 1 = 6 * 6 = 36 . . ways of selecting all three = 4 c 3 = 4 = 4 . . total = 60 + 36 + 4 = 100 . . . answer : a