The question is in the picture ! idk how to do this

35

Step-by-step explanation:

Put the numbers in the formula and do the arithmetic.

nCk = n!/(k!(n -k)!)

7C3 = 7!/(3!(7-3)!) = 7·6·5/(3·2·1) = 7·5 = 35

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It is convenient to use the largest of the factorials in the denominator to cancel as many factors as you can from the numerator, then cancel factors from the remaining numbers. Here after canceling 4! = 4·3·2·1 from the numerator, we are left with 7·6·5 divided by 3! = 3·2·1 = 6. Obviously, this will cancel the 6 in the numerator product, leaving only 7·5 = 35.

Some graphing and/or scientific calculators will have this function built in.

21/58.

hope to be marked brainliest