₄₉C₅ = the number of ways a group of 5 items can be chosen from a selection of 49. As a group it does not matter in which order the 5 are selected.
It is calculated as:
₄₉C₅ = 49! / ((49-5)! 5!) = 49! / (44! 5!) = 49 × 48 × 47 × 46 × 45 / 5 × 4 × 3 × 2 × 1 = 1,906,884
(The font used to display Answers has a very bad exclamation mark (!) which looks like a vertical bar. 49! is 49 factorial and 49! = 49 × 48 × 47 × ... × 3 × 2 × 1)
There are only 5 distinct combinations of 4 numbers.(1234, 1235, 1245, 1345, 2345) C = 5! / 4!But there are 120 distinct combinations in distinct order (i.e. 24 ways to order each abcd).abcdabdcacbdacdbadbcadcb
To find the number of combinations of 3 numbers from a set of 5 numbers, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ). Here, ( n = 5 ) and ( k = 3 ). Plugging in the values, we get ( C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 ). Therefore, there are 10 combinations of 3 numbers from a set of 5 numbers.
81 deg F Subtract 32: 81 - 32 = 49 Multiply by 5: 49*5 = 245 Divide by 9: 245/9 = 27.22... deg C
#include<stdio.h> int main() { int a,b,c,d; for(a=1; a<5; a++) { for(b=1; b<5; b++) { for(c=1; c<5; c++) { for(d=1; d<5; d++) { if(!(a==b a==c a==d b==c b==d c==d)) printf("dd\n",a,b,c,d); } } } } return 0; }
Suppose the 5 letters are A, B, C, D and E. The letter A can either be in the combination or not: 2 options for A. With each of these options, B can either be in the combination or not: 2 options for B - making 2*2 options so far. With each of the options so far, C can either be in the combination or not: 2 options for C - making 2*2*2 options so far. and so on. So for 5 letters there are 25 = 32 combinations. However, one of these is the combination that excludes each of the 5 letters - ie the null combination. Excluding the null combination gives the final answer of 31 combinations.
If the question concerned the number of combinations of three different coins, the answer is 23-1 = 7. If the coins are a,b,and c, the combinations are a, b, c, ab, ac, bc, abc. If two of the coins are the same there are only 5 combinations and if all three are the same there are 3.
c + c + 1 = 49 2*c + 1 = 492*c = 48c = 24c + c + 1 = 49 2*c + 1 = 492*c = 48c = 24c + c + 1 = 49 2*c + 1 = 492*c = 48c = 24c + c + 1 = 49 2*c + 1 = 492*c = 48c = 24
There are only 5 distinct combinations of 4 numbers.(1234, 1235, 1245, 1345, 2345) C = 5! / 4!But there are 120 distinct combinations in distinct order (i.e. 24 ways to order each abcd).abcdabdcacbdacdbadbcadcb
The pass rate for GCSE Computing is a minimum of grade C.
Lux Video Theatre - 1950 The Bride Came C-O-D- 5-49 was released on: USA: 28 July 1955
C++, JAVA, VB.NET, Python, Fortran, PHP and HTML 5
5 digit combinations for numbers 1 to 50= 50 C 5N!= 1x2x3x4x...x (N-1) x N= (50!)/(45!x5!) =(50x49x...x2x1)/((45x44x...x2x1)(5x4x3x2x1))=(50x49x48x47x46)/(5x4x3x2)=254251200/120 =21187602118760 combinations
C 5 2 = 5! / (2! x 3!) = (5x4x3x2x1) / (2x1 x 3x2x1) = 10
81 deg F Subtract 32: 81 - 32 = 49 Multiply by 5: 49*5 = 245 Divide by 9: 245/9 = 27.22... deg C
They are all programming languages.
C
c is platform dependent