70C4
= 70!/[(70 - 4)!4!]
= 70!/(66!4!)
= (70 x 69 x 68 x 67 x 66!)/(66! x 4 x 3 x 2 x 1)
= 35 x 23 x 17 x 67
= 916,895
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To find the number of different combinations of 10 numbers from a total of 70 numbers, you can use the combination formula, which is represented as ( C(n, r) = \frac{n!}{r!(n-r)!} ). In this case, ( n = 70 ) and ( r = 10 ), so the calculation is ( C(70, 10) = \frac{70!}{10!(70-10)!} ). This results in a total of 5,486,560 different combinations.
There are 10,000 four number combinations in the numbers 0 through 9. * * * * * No, that is simply not true. The combination 1234 is the same as the combination 2134 and 2314 and 1423 etc. So, there are just 8*7*6*5/(4*3*2*1) = 70 combinations.
The number of combinations of 4 numbers out of 8 is 8C4= (8*7*6*5/(4*3*2*1) = 70.In a combination, the order of the numbers does not matter so that 2345, 2454, 4523 etc are all counted as 1.The number of combinations of 4 numbers out of 8 is 8C4= (8*7*6*5/(4*3*2*1) = 70.In a combination, the order of the numbers does not matter so that 2345, 2454, 4523 etc are all counted as 1.The number of combinations of 4 numbers out of 8 is 8C4= (8*7*6*5/(4*3*2*1) = 70.In a combination, the order of the numbers does not matter so that 2345, 2454, 4523 etc are all counted as 1.The number of combinations of 4 numbers out of 8 is 8C4= (8*7*6*5/(4*3*2*1) = 70.In a combination, the order of the numbers does not matter so that 2345, 2454, 4523 etc are all counted as 1.
75*74*73*72*71*70*69*68/(8*7*6*5*4*3*2*1) = 16,871,053,725
The two numbers that have a sum of 70 are 35 and 35. This is because when you add 35 and 35 together, you get 70. Another way to think about it is that if you have two numbers x and y that add up to 70, then x + y = 70. By substituting 35 for both x and y, you get 35 + 35 = 70.