There are 28C6 = (28*27*26*25*24*23)/(6*5*4*3*2*1) = 376,740 sets.
-99
Answer is 19. Explanation: we look at the entire range 99 to 131, in smaller groups. 99 has no digit 1. 100 to 109 have nine numbers with only one "1". 110 to 119 have no numbers with only one "1". 120 to 129 have nine numbers with only one "1". 130 has one number with only one "1". 131 does not have only one "1". Total: 9 + 9 + 1 = 19
There are 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 combinations.
25 numbers are between 1-100.
Infinitely many. In fact there are more irrational numbers between 1 and 10 than there are rational numbers - in total!
Yes, they can.
There are 4 groups of 2 in 8 (2X4=8, 8/2=4)There are 28 pairs in all the numbers between 1 and 8 (1&2,1&3, etc.)
The answer is 8C6 = 8!/[6!*(8-6)!] where n! = 1*2*3*...*n So, the answer is 28.
1, 2, 4, 7, 14, 28.
1, 2, 4, 7, 14, and 28 are the numbers that match this description. Therefore, six numbers divide into it evenly.
If the six numbers are all different, then the answer is 6C4 = 6*5/(2*1) = 15
There are infinitely many possible answers: 1*2*14 = 28 or 5*20*0.28 = 28
56 will split into the following groups... 56 groups of 1 28 groups of 2 14 groups of 4 7 groups of 8 8 groups of 7 4 groups of 14 2 groups of 28 and 1 group of 56
It looks like you are asking how many combinations of 6 numbers are there in the 28 numbers 1 through 28. This is known as the number of combinations of 28 things taken 6 at a time. The answer is 28!/(6!22!) (n! means n factorial, which is the product of all the integers from 1 to n). I get 376,740 if I haven't made an error in arithmetic.
Two numbers that have an LCM of 28 are 14 and 28.
There are 8*7/(2*1) = 28 combinations.
1, 2, 4, 7, 14, and 28 are the numbers you seek.