Combination worksheet

1. Combinations with repetition: 3,162,510

1. Items can be reused, so count it as choosing r from n + r − 1 (the stars-and-bars method)

   $C\left(n+r-1,r\right)$$=\frac{\left(n+r-1\right)\text{!}}{r\text{!}\left(n-1\right)\text{!}}$

2. Here n + r − 1 = 50 + 5 − 1 = 54

   $50+5-1$$=54$

3. Substitute into the factorial formula: 54! divided by 5! times 49!

   $C\left(54,5\right)$$=\frac{54\text{!}}{5\text{!}\times 49\text{!}}$

4. Cancel 49! from the top and bottom. What is left is the top 5 factors of 54! divided by 5!

   $\frac{54\times 53\times 52\times 51\times 50}{5\text{!}}$$=\frac{\mathrm{379,501,200}}{120}$

5. Divide to get the number of combinations

   $C\left(54,5\right)$$=\mathrm{3,162,510}$

2. Combinations with repetition: 178,365

1. Items can be reused, so count it as choosing r from n + r − 1 (the stars-and-bars method)

   $C\left(n+r-1,r\right)$$=\frac{\left(n+r-1\right)\text{!}}{r\text{!}\left(n-1\right)\text{!}}$

2. Here n + r − 1 = 44 + 4 − 1 = 47

   $44+4-1$$=47$

3. Substitute into the factorial formula: 47! divided by 4! times 43!

   $C\left(47,4\right)$$=\frac{47\text{!}}{4\text{!}\times 43\text{!}}$

4. Cancel 43! from the top and bottom. What is left is the top 4 factors of 47! divided by 4!

   $\frac{47\times 46\times 45\times 44}{4\text{!}}$$=\frac{\mathrm{4,280,760}}{24}$

5. Divide to get the number of combinations

   $C\left(47,4\right)$$=\mathrm{178,365}$

3. Combinations, C(n, r): 12,271,512

1. Order doesn't matter and no item is reused, so use the combinations formula

   $C\left(n,r\right)$$=\frac{n\text{!}}{r\text{!}\left(n-r\right)\text{!}}$

2. Substitute into the factorial formula: 48! divided by 6! times 42!

   $C\left(48,6\right)$$=\frac{48\text{!}}{6\text{!}\times 42\text{!}}$

3. Cancel 42! from the top and bottom. What is left is the top 6 factors of 48! divided by 6!

   $\frac{48\times 47\times 46\times 45\times 44\times 43}{6\text{!}}$$=\frac{\mathrm{8,835,488,640}}{720}$

4. Divide to get the number of combinations

   $C\left(48,6\right)$$=\mathrm{12,271,512}$

4. Combinations, C(n, r): 61,474,519

1. Order doesn't matter and no item is reused, so use the combinations formula

   $C\left(n,r\right)$$=\frac{n\text{!}}{r\text{!}\left(n-r\right)\text{!}}$

2. Substitute into the factorial formula: 62! divided by 6! times 56!

   $C\left(62,6\right)$$=\frac{62\text{!}}{6\text{!}\times 56\text{!}}$

3. Cancel 56! from the top and bottom. What is left is the top 6 factors of 62! divided by 6!

   $\frac{62\times 61\times 60\times 59\times 58\times 57}{6\text{!}}$$=\frac{\mathrm{44,261,653,680}}{720}$

4. Divide to get the number of combinations

   $C\left(62,6\right)$$=\mathrm{61,474,519}$

5. Combinations, C(n, r): 194,580

1. Order doesn't matter and no item is reused, so use the combinations formula

   $C\left(n,r\right)$$=\frac{n\text{!}}{r\text{!}\left(n-r\right)\text{!}}$

2. Substitute into the factorial formula: 48! divided by 4! times 44!

   $C\left(48,4\right)$$=\frac{48\text{!}}{4\text{!}\times 44\text{!}}$

3. Cancel 44! from the top and bottom. What is left is the top 4 factors of 48! divided by 4!

   $\frac{48\times 47\times 46\times 45}{4\text{!}}$$=\frac{\mathrm{4,669,920}}{24}$

4. Divide to get the number of combinations

   $C\left(48,4\right)$$=\mathrm{194,580}$

6. Combinations, C(n, r): 3,838,380

1. Order doesn't matter and no item is reused, so use the combinations formula

   $C\left(n,r\right)$$=\frac{n\text{!}}{r\text{!}\left(n-r\right)\text{!}}$

2. Substitute into the factorial formula: 40! divided by 6! times 34!

   $C\left(40,6\right)$$=\frac{40\text{!}}{6\text{!}\times 34\text{!}}$

3. Cancel 34! from the top and bottom. What is left is the top 6 factors of 40! divided by 6!

   $\frac{40\times 39\times 38\times 37\times 36\times 35}{6\text{!}}$$=\frac{\mathrm{2,763,633,600}}{720}$

4. Divide to get the number of combinations

   $C\left(40,6\right)$$=\mathrm{3,838,380}$

7. Combinations, C(n, r): 1,984,829,850

1. Order doesn't matter and no item is reused, so use the combinations formula

   $C\left(n,r\right)$$=\frac{n\text{!}}{r\text{!}\left(n-r\right)\text{!}}$

2. Substitute into the factorial formula: 75! divided by 7! times 68!

   $C\left(75,7\right)$$=\frac{75\text{!}}{7\text{!}\times 68\text{!}}$

3. Cancel 68! from the top and bottom. What is left is the top 7 factors of 75! divided by 7!

   $\frac{75\times 74\times 73\times 72\times 71\times 70\times 69}{7\text{!}}$$=\frac{\mathrm{10,003,542,444,000}}{\mathrm{5,040}}$

4. Divide to get the number of combinations

   $C\left(75,7\right)$$=\mathrm{1,984,829,850}$

8. Combinations, C(n, r): 82,598,880

1. Order doesn't matter and no item is reused, so use the combinations formula

   $C\left(n,r\right)$$=\frac{n\text{!}}{r\text{!}\left(n-r\right)\text{!}}$

2. Substitute into the factorial formula: 65! divided by 6! times 59!

   $C\left(65,6\right)$$=\frac{65\text{!}}{6\text{!}\times 59\text{!}}$

3. Cancel 59! from the top and bottom. What is left is the top 6 factors of 65! divided by 6!

   $\frac{65\times 64\times 63\times 62\times 61\times 60}{6\text{!}}$$=\frac{\mathrm{59,471,193,600}}{720}$

4. Divide to get the number of combinations

   $C\left(65,6\right)$$=\mathrm{82,598,880}$

9. Combinations with repetition: 249,900

1. Items can be reused, so count it as choosing r from n + r − 1 (the stars-and-bars method)

   $C\left(n+r-1,r\right)$$=\frac{\left(n+r-1\right)\text{!}}{r\text{!}\left(n-1\right)\text{!}}$

2. Here n + r − 1 = 48 + 4 − 1 = 51

   $48+4-1$$=51$

3. Substitute into the factorial formula: 51! divided by 4! times 47!

   $C\left(51,4\right)$$=\frac{51\text{!}}{4\text{!}\times 47\text{!}}$

4. Cancel 47! from the top and bottom. What is left is the top 4 factors of 51! divided by 4!

   $\frac{51\times 50\times 49\times 48}{4\text{!}}$$=\frac{\mathrm{5,997,600}}{24}$

5. Divide to get the number of combinations

   $C\left(51,4\right)$$=\mathrm{249,900}$

10. Combinations with repetition: 292,825

1. Items can be reused, so count it as choosing r from n + r − 1 (the stars-and-bars method)

   $C\left(n+r-1,r\right)$$=\frac{\left(n+r-1\right)\text{!}}{r\text{!}\left(n-1\right)\text{!}}$

2. Here n + r − 1 = 50 + 4 − 1 = 53

   $50+4-1$$=53$

3. Substitute into the factorial formula: 53! divided by 4! times 49!

   $C\left(53,4\right)$$=\frac{53\text{!}}{4\text{!}\times 49\text{!}}$

4. Cancel 49! from the top and bottom. What is left is the top 4 factors of 53! divided by 4!

   $\frac{53\times 52\times 51\times 50}{4\text{!}}$$=\frac{\mathrm{7,027,800}}{24}$

5. Divide to get the number of combinations

   $C\left(53,4\right)$$=\mathrm{292,825}$