Permutation combination practice questions

One key thing to keep in mind with combinatorics is that there are very few formulae to remember. It's more about logical thinking and enumerating all the cases correctly.


Q1 - How many Arithmetic Progressions can be formed if first term is a 2 digit no. and common diff is a single digit no.(non-zero).
Answer: 90 * 9.

Q2- If 3 numbers are selected from the set of first 100 natural nos such that their sum is divisible by 3; then find the no. of ways in which it can be done.

Answer:
Case 1: All 3 nums with remainder 0 (i.e. all divisible by 3): \(\binom{33}{3}\)
Case 2: All 3 nums with remainder 1: \(\binom{34}{3}\)
Case 3: All 3 nums with remainder 2: \(\binom{33}{3}\)
Case 4: 1 number each with reaminder 0,1,2: \(\binom{33}{1}\) * \(\binom{34}{1}\) * \(\binom{33}{1}\)


Q3- A box contains 5 diff red & 6 diff white balls. In how many ways can 6 balls be selected so that there are atleast 2 red balls.

Answer:
Total ways to select: \(\binom{11}{6}\)
Now subtract the cases when all are white or only 1 is white: \(\binom{6}{6}\) * \(\binom{5}{0}\) + \(\binom{6}{5}\) * \(\binom{5}{1}\).
So finally: \(\binom{11}{6}\) - \(\binom{6}{6}\) * \(\binom{5}{0}\) - \(\binom{6}{5}\) * \(\binom{5}{1}\) = 431.

Q4 - Find the no. of ways of selecting 5 members from a group of 7 men & 5 women such that
a) No restriction
Answer: \(\binom{12}{5}\)

b) Men in majority
Answer: \(\binom{7}{3}\) * \(\binom{5}{2}\) + \(\binom{7}{4}\) * \(\binom{5}{1}\) + \(\binom{7}{5}\) * \(\binom{5}{0}\)

c) at least 2 men and 1 woman
Answer: \(\binom{7}{2}\) * \(\binom{5}{3}\) + \(\binom{7}{3}\) * \(\binom{5}{2}\) + \(\binom{7}{4}\) * \(\binom{5}{1}\)

Q5- Out of 16 players of a cricket team, 4 are bowlers and 2 are wicket keepers & remaining batsmen. A team of 11 players is to chosen so as to contain atleast 3 bowlers & 1 wicket keeper. Find the no. of ways.

Answer:
\(\binom{4}{3}\) * \(\binom{2}{1}\) * \(\binom{10}{7}\) +
\(\binom{4}{3}\) * \(\binom{2}{2}\) * \(\binom{10}{6}\) +
\(\binom{4}{4}\) * \(\binom{2}{1}\) * \(\binom{10}{6}\) +
\(\binom{4}{4}\) * \(\binom{2}{2}\) * \(\binom{10}{5}\) +

Q6 – A delegation of 4 students is to be selected from a total of 12 students. In how many ways this can be done if

a) all the students are equally willing
Answer: \(\binom{12}{4}\)

b) 2 particular students have to be included
Answer: \(\binom{10}{2}\)

c) 2 particular students can't be selected together
Answer:
\(\binom{10}{4}\) + \(\binom{11}{3}\) + \(\binom{11}{3}\)
Another way to write the same thing: \(\binom{12}{4}\) + \(\binom{10}{2}\)

Q7 – Find the total number of signals that can be made by 5 different flags.

Answer: \(\binom{5}{1}\) * 1! + \(\binom{5}{2}\) * 2! + \(\binom{5}{3}\) * 3! + \(\binom{5}{4}\) * 4! + \(\binom{5}{5}\) * 5!

Q8 – From 6 different novels & 3 different dictionaries, 4 novels & 1 dictionary are to be selected & arranged in a row so that dictionary is always in middle. Find the number of ways.
Answer: \(\binom{6}{4}\) * \(\binom{3}{1}\) *4!

Q9: How many 4-digit no. can be formed such that (no repetition, non-zero digits)
a) No restrictions
Answer: 9*8*7*6

b) No is even
Answer: 4*8*7*6

c) No is odd
5*8*7*6

d) divisible by 5
8*7*6

e) at least one digit is even
Answer: From total numbers, subtract the ones with no even digit.
9*8*7*6 - 5*4*3*2

f) last 2 digits are twin prime
So last 2 digits can be 3,5 or 5,3 or 5,7 or 7,5.
Rest of the digits can be chosen in 7*6 ways.
Finally: 4*7*6.

Q10: How many 4-digit no. can be formed such that (no repetition)

Now we have to consider 0 also.
a) No restriction
9*9*8*7

b) No is even
Case 1: Last digit is 0: 9*8*7.
Case 2: Last digit is 2,4,6,8: 4*8*8*7.
Add both: 9*8*7 + 4*8*8*7.

c) No is odd
5*8*8*7

d) No is divisible by 5
Case 1: Last digit is 0. 9*8*7.
Case 2: Last digit is 5. 8*8*7.
Total : 9*8*7 + 8*8*7.

Q11. Find the no. of elements in the set \(X = \{x; \, x \in \mathbb{Q}, \, 0 < x < 1, \, x = \frac{p}{q} \}\) where p, q are chosen from the set \(\{1, 2, 3, 4, 5, 6\}\)

Answer: 11
1/2,1/3,1/4,1/5,1/6,2/3,2/5,3/4,3/5,4/5,5/6

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