Permutations and Combinations: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 4)

The word ‘GARDEN’ has 6 letters, and all are distinct. Since every letter is different, each position matters for the order. Number of arrangements = 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720. So ‘GARDEN’ can be arranged in 720 ways.

The word ‘MARKET’ has 6 letters, and all are distinct. Since every letter is different, each position matters for the order. Number of arrangements = 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720. So ‘MARKET’ can be arranged in 720 ways.

The word ‘SILVER’ has 6 letters, and all are distinct. Since every letter is different, each position matters for the order. Number of arrangements = 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720. So ‘SILVER’ can be arranged in 720 ways.

We compare 28C21 and 28C7. Choosing 21 items to include from 28 automatically leaves 7 items excluded. So choosing 21 items to include is equivalent to choosing 7 items to leave out. This is the identity nCr = nC(n-r): 28C21 = 28C7.

We compare 29C27 and 29C2. Choosing 27 items to include from 29 automatically leaves 2 items excluded. So choosing 27 items to include is equivalent to choosing 2 items to leave out. This is the identity nCr = nC(n-r): 29C27 = 29C2.

We need 23C23, choosing all 23 items out of 23. There is only one way to choose every single item: take them all. So 23C23 = 1.

We need 30C30, choosing all 30 items out of 30. There is only one way to choose every single item: take them all. So 30C30 = 1.

We need 5C5, choosing all 5 items out of 5. There is only one way to choose every single item: take them all. So 5C5 = 1.

We need 9C0, choosing 0 items out of 9. There is only one way to choose nothing: the empty selection. So 9C0 = 1.

We need 5C0, choosing 0 items out of 5. There is only one way to choose nothing: the empty selection. So 5C0 = 1.

We compare 20C18 and 20C2. Choosing 18 items to include from 20 automatically leaves 2 items excluded. So choosing 18 items to include is equivalent to choosing 2 items to leave out. This is the identity nCr = nC(n-r): 20C18 = 20C2.

We need 18C5, combinations of 18 items taken 5 at a time. We’re choosing 5 items out of 18, where order does NOT matter, so we use combinations. nCr = n! / (r!(n-r)!), so 18C5 = 18! / (5! x 13!). This equals (18 x 17 x 16 x 15 x 14) / 5! = (18 x 17 x 16 x 15 x 14) / 120 = 8568. So 18C5 = 8568.

We need 18C6, combinations of 18 items taken 6 at a time. We’re choosing 6 items out of 18, where order does NOT matter, so we use combinations. nCr = n! / (r!(n-r)!), so 18C6 = 18! / (6! x 12!). This equals (18 x 17 x 16 x 15 x 14 x 13) / 6! = (18 x 17 x 16 x 15 x 14 x 13) / 720 = 18564. So 18C6 = 18564.

We need 16C4, combinations of 16 items taken 4 at a time. We’re choosing 4 items out of 16, where order does NOT matter, so we use combinations. nCr = n! / (r!(n-r)!), so 16C4 = 16! / (4! x 12!). This equals (16 x 15 x 14 x 13) / 4! = (16 x 15 x 14 x 13) / 24 = 1820. So 16C4 = 1820.

We need 20C5, combinations of 20 items taken 5 at a time. We’re choosing 5 items out of 20, where order does NOT matter, so we use combinations. nCr = n! / (r!(n-r)!), so 20C5 = 20! / (5! x 15!). This equals (20 x 19 x 18 x 17 x 16) / 5! = (20 x 19 x 18 x 17 x 16) / 120 = 15504. So 20C5 = 15504.

We need 17C2, combinations of 17 items taken 2 at a time. We’re choosing 2 items out of 17, where order does NOT matter, so we use combinations. nCr = n! / (r!(n-r)!), so 17C2 = 17! / (2! x 15!). This equals (17 x 16) / 2! = (17 x 16) / 2 = 136. So 17C2 = 136.

We need 12C2, combinations of 12 items taken 2 at a time. We’re choosing 2 items out of 12, where order does NOT matter, so we use combinations. nCr = n! / (r!(n-r)!), so 12C2 = 12! / (2! x 10!). This equals (12 x 11) / 2! = (12 x 11) / 2 = 66. So 12C2 = 66.

We need 11C6, combinations of 11 items taken 6 at a time. We’re choosing 6 items out of 11, where order does NOT matter, so we use combinations. nCr = n! / (r!(n-r)!), so 11C6 = 11! / (6! x 5!). This equals (11 x 10 x 9 x 8 x 7 x 6) / 6! = (11 x 10 x 9 x 8 x 7 x 6) / 720 = 462. So 11C6 = 462.

We need 8C4, combinations of 8 items taken 4 at a time. We’re choosing 4 items out of 8, where order does NOT matter, so we use combinations. nCr = n! / (r!(n-r)!), so 8C4 = 8! / (4! x 4!). This equals (8 x 7 x 6 x 5) / 4! = (8 x 7 x 6 x 5) / 24 = 70. So 8C4 = 70.

We need 8C2, combinations of 8 items taken 2 at a time. We’re choosing 2 items out of 8, where order does NOT matter, so we use combinations. nCr = n! / (r!(n-r)!), so 8C2 = 8! / (2! x 6!). This equals (8 x 7) / 2! = (8 x 7) / 2 = 28. So 8C2 = 28.

We need 4C3, combinations of 4 items taken 3 at a time. We’re choosing 3 items out of 4, where order does NOT matter, so we use combinations. nCr = n! / (r!(n-r)!), so 4C3 = 4! / (3! x 1!). This equals (4 x 3 x 2) / 3! = (4 x 3 x 2) / 6 = 4. So 4C3 = 4.

We need 7P3, permutations of 7 items taken 3 at a time. We’re arranging 3 items out of 7, where order matters, so we use permutations. nPr = n! / (n-r)!, so 7P3 = 7! / 4!. This is the product of 3 numbers counting down from 7: 7 x 6 x 5 = 210. So 7P3 = 210.

We need 6P5, permutations of 6 items taken 5 at a time. We’re arranging 5 items out of 6, where order matters, so we use permutations. nPr = n! / (n-r)!, so 6P5 = 6! / 1!. This is the product of 5 numbers counting down from 6: 6 x 5 x 4 x 3 x 2 = 720. So 6P5 = 720.

We need 15P5, permutations of 15 items taken 5 at a time. We’re arranging 5 items out of 15, where order matters, so we use permutations. nPr = n! / (n-r)!, so 15P5 = 15! / 10!. This is the product of 5 numbers counting down from 15: 15 x 14 x 13 x 12 x 11 = 360360. So 15P5 = 360360.

We need 6P4, permutations of 6 items taken 4 at a time. We’re arranging 4 items out of 6, where order matters, so we use permutations. nPr = n! / (n-r)!, so 6P4 = 6! / 2!. This is the product of 4 numbers counting down from 6: 6 x 5 x 4 x 3 = 360. So 6P4 = 360.