Questions: Permutations and Combinations

Q1. A 3-digit number is built from the digits 1, 2, 3, 4 and 5 with no digit repeated. Count the possibilities.

Answer: The first place has 5 choices. Once a digit is used, the second place has 4 choices and the third has 3. Multiplying gives 5 times 4 times 3 equal to 60. This is also 5P3. A quick check on the count: the largest possible value is 543 and the smallest is 123, and 60 numbers fit inside that range once order is counted.

Hinglish: Pehli jagah 5 choices, phir 4, phir 3. Product 5 x 4 x 3 = 60, yaani 5P3. Digits repeat nahi hote, isliye choices ghatti jaati hain.

Q2. A customer must choose a shirt from 4 options, a pair of trousers from 3 options and a pair of shoes from 2 options. How many outfits can be formed?

Answer: The choices form stages, so multiply: 4 times 3 times 2 equal to 24 outfits. No divisibility by any factorial is needed because each choice is independent and no item repeats. Notice that changing the order in which the three decisions are made does not change the product, which is the commutative property at work inside the counting principle.

Hinglish: Stages ke choices multiply karo: 4 x 3 x 2 = 24 outfits. Har choice independent hai, isliye factorial se divide nahi karna.

Q3. Evaluate 8 factorial divided by 6 factorial.

Answer: Expand the numerator up to the smaller factorial: 8 factorial is 8 times 7 times 6 factorial. Cancelling 6 factorial leaves 8 times 7 equal to 56. Expanding fully would give 40320 by 720, the same value but more work.

Hinglish: 8! = 8 x 7 x 6! likho. 6! cancel karke 8 x 7 = 56 bachta hai. Poora expand karne ki zaroorat nahi.

Q4. The four letters of ROSE are used once each to make a word. How many words are possible?

Answer: All four letters are different, so the number of arrangements is 4 factorial equal to 24. Each arrangement uses every letter exactly once, so no denominator is needed. A useful check is that each of the four letters can be placed first, and the remaining three can be arranged in 6 ways, giving 4 times 6 equal to 24.

Hinglish: Chaar letters different hain, isliye arrangements 4! = 24. Har letter pehle aa sakta hai aur baaki teen 3! = 6 tarah se arrange hote hain, to 4 x 6 = 24.

Q5. Arrange the letters of BANANA in distinct ways.

Answer: Six letters are present, with A three times and N twice. Divide the total arrangements by the factorials of the repeated counts: 6 factorial divided by 3 factorial times 2 factorial, equal to 720 by 12, that is 60 distinct arrangements. Dividing is needed because swapping two A's produces the same word, and so does swapping two N's.

Hinglish: Chhe letters hain, A teen baar aur N do baar. 6!/(3! x 2!) = 720/12 = 60 distinct arrangements. Do A ya do N swap karne se wahi word banta hai, isliye divide karte hain.

Q6. In how many ways can a committee of 3 be chosen from 10 people?

Answer: A committee is a selection, so order does not matter. The count is 10C3, which equals 10 factorial by 3 factorial times 7 factorial, that is 120. Choosing the same three people in a different order gives the same committee and is not counted again. A way to see the arithmetic: 10 times 9 times 8 is 720 arrangements of three distinct people, and each committee appears in 3 factorial equal to 6 orders, so divide 720 by 6.

Hinglish: Committee me order matter nahi karta, isliye 10C3 = 120. Same logon ka alag order same committee hai.

Q7. From 5 boys and 4 girls, how many teams of 2 boys and 2 girls can be formed?

Answer: Choose the boys in 5C2 equal to 10 ways and the girls in 4C2 equal to 6 ways. The choices are independent, so multiply: 10 times 6 equal to 60 teams.

Hinglish: Boys 5C2 = 10, girls 4C2 = 6. Independent choices hain, isliye 10 x 6 = 60 teams.

Q8. An even 4-digit number is built from 1, 2, 3, 4 and 5 with all digits different. Count the numbers.

Answer: For an even number the last digit is 2 or 4, giving 2 choices. The first three places are then filled from the remaining four digits in 4P3 equal to 24 ways. Multiplying gives 2 times 24 equal to 48 numbers.

Hinglish: Even number ke liye last digit 2 ya 4, yaani 2 choices. Baaki teen places 4P3 = 24 tarah se bharte hain. Total 2 x 24 = 48.

Q9. How many ways can 5 people be seated in a row if two particular people must sit together?

Answer: Treat the pair as one block. The block and the other three people make 4 objects, which can be arranged in 4 factorial equal to 24 ways. Inside the block the pair can swap in 2 factorial equal to 2 ways. The total is 24 times 2 equal to 48.

Hinglish: Do logon ko ek block maano. Block plus baaki teen log = 4 objects, unke 4! = 24 arrangements. Block ke andar pair 2! = 2 tarah se baithe. Total 24 x 2 = 48.

Q10. Nine seats in a row are filled so that the four women take the even positions. Count the arrangements when five men take the remaining seats.

Answer: A row of 9 seats has four even places and five odd places. The four women take the four even places in 4 factorial equal to 24 ways. The five men fill the five odd places in 5 factorial equal to 120 ways. The choices are independent, so the total is 24 times 120 equal to 2880 arrangements.

Hinglish: 9 seats me 4 even aur 5 odd places hain. Chaar women 4 even places par 4! = 24 tarah se baith sakti hain, aur paanch men 5 odd places par 5! = 120 tarah se. Total 24 x 120 = 2880.

Common mistakes on this chapter

Using nPr where the question needs a selection is the first error. Forgetting repeated letters in a word is the second. Counting at least cases by fixing one required item first is the third, because it double counts. Ignoring a leading zero restriction in a number question is the fourth. Read every answer once with these four checks in mind.

Quiz

Quiz below: five multiple choice questions on the counting principle, factorials, permutations, combinations and restricted arrangements. Attempt them after the notes above, then move to the previous year page.