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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)

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Q: What is 49 C 5 in computing combinations?

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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

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.

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

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

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++, JAVA, VB.NET, Python, Fortran, PHP and HTML 5

#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; }

C 5 2 = 5! / (2! x 3!) = (5x4x3x2x1) / (2x1 x 3x2x1) = 10

There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.

There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.There are an infinite number of combinations for A, B, and C in this case. Just assign any number for A and for B, then calculate the value for C.

They are all programming languages.

81 deg F Subtract 32: 81 - 32 = 49 Multiply by 5: 49*5 = 245 Divide by 9: 245/9 = 27.22... deg C

3125 i think...There are 5 numbers.(It can be any five, say 9,7,6,2,1)To write them in different combinations you will do it in n! (factorial n) number of ways.n! = 5! = 5x4x3x2x1 = 120 different combinations.Even if the numbers were replaced by say A,B,C,D,E (or any 5 different letters or numbers) then it will still be 120 combinations for those 5 letters or numbers.

49 degrees Fahrenheit = 9.4 degrees CelsiusConversion: [°C] = ([°F] − 32) × 5⁄9

I think you're trying to ask: In what environment does the Fahrenheit thermometerread a number that is 49 higher than the reading on the Celsius thermometer ?F = (1.8 C) + 32andF = C + 49soC + 49 = 1.8 C + 32C + 17 = 1.8 C17 = 0.8 CC = 17/0.8C = 21.25Â°F = 70.25Â°Check:==> 70.25 - 21.25 = 49==> 1.8 C + 32 = (1.8) (21.25) + 32 = 38.25 + 32 = 70.25 yay!

c is platform dependent

(3c +15)/(c2 - 5) + c/(c + 5)= (3c + 15)/(c - 5)(c + 5) + c/(c + 5)= (3c + 15)/(c - 5)(c + 5) + [c(c - 5)/(c + 5)(c - 5)(since the common denominator is (c + 5)(c - 5)= (3c + 15)/(c - 5)(c + 5) + (c2 - 5c)/(c + 5)(c - 5)= (3c + 15 + c2 - 5c)/(c - 5)(c + 5)= (c2 - 2c + 15)/(c - 5)(c + 5)= [(c - 5)(c + 3)]/(c - 5)(c + 5) (simplify)= (c + 3)/(c + 5)

#include <stdio.h> int main(int argc, char** argv){ 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("%d%d%d%d\n",a,b,c,d); } } } } return 0; }

49 cups is 392 fluid ounces.

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.

Centre for devlopment of advanced computing.

49

cry on me49

Recombination, or crossing over, is the exchanging of DNA between homologous chromosomes at meiosis. It results in novel combinations of alleles in the gametes, that is, it scrambles the alleles into new combinations, Let's take an example of an individual that has genes A,B and C on one chromsome, and has the alleles a and c on the other chromosome of the pair: A---B---C a---B---c If crossing over occurs between the genes A and B (and not between B and C), then we can get the following genotypes in the gametes, in addition to the above two combinations: : A---B---c a---B---C Crossing over has scrambled the allelic combinations and given us two new additional ones in the gametes. If we allow recombination between B and C, then even more combinations are possible. One can easily see how recombination can add enormous amounts of genetic variation into a population. Also, note that if no recombination occurs, only the two original combinations of alleles will make it into the gametes.