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TISSNET Permutations and Combinations — practice questions

Explore 11 free Permutations and Combinations questions in TISSNET Quantitative Aptitude. Open each question or exercise to read its content, reference answers and available explanation.

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Permutations and Combinations questions and exercises with answers

Showing questions 1–11 of 11.

The factorial of a non-negative integer $n$, denoted $n!$, is defined asRead the answer and worked solution →Find the number of arrangements of all the letters in the word 'CAT'.Read the answer and worked solution →Find the number of distinct arrangements of all the letters in the word 'MISSISSIPPI'.Read the answer and worked solution →How many ways can 5 people be seated around a CIRCULAR table?Read the answer and worked solution →From a group of 8 people, how many ways can a COMMITTEE of 3 be selected?Read the answer and worked solution →In how many ways can the letters of the word 'TRIANGLE' be arranged such that all the VOWELS are TOGETHER?Read the answer and worked solution →How many distinct 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 (each digit can be used at MOST once)?Read the answer and worked solution →The number of ways to select 2 boys and 3 girls from a group of 5 boys and 4 girls isRead the answer and worked solution →Find the number of ways a person can travel from city A to city C if there are 3 routes from A to B and 2 routes from B to C.Read the answer and worked solution →$\binom{n}{r} + \binom{n}{r-1}$ equalsRead the answer and worked solution →How many DIAGONALS does a regular DECAGON (10-sided polygon) have?Read the answer and worked solution →

Worked practice questions

The factorial of a non-negative integer $n$, denoted $n!$, is defined as

  1. $n \cdot (n+1) \cdot (n+2) \cdots$
  2. $n \cdot n$
  3. $n + (n-1) + (n-2) + \ldots + 1$
  4. $1 \cdot 2 \cdot 3 \cdots n$, with $0! = 1$
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Correct answer: D. $1 \cdot 2 \cdot 3 \cdots n$, with $0! = 1$

1. Your syllabus: $n!$ is the PRODUCT of all positive integers from 1 to $n$ inclusive. 2. $n! = 1 \cdot 2 \cdot 3 \cdots n$. 3. By CONVENTION: $0! = 1$ (this makes formulas with $n!$ work consistently when $n = 0$). 4. Examples: $5! = 120$, $10! = 3{,}628{,}800$. 5. RECURSIVE definition: $n! = n \cdot (n-1)!$, which gives the same values. 6. Option A goes upward (infinite product). Option C ignores the product over all integers. Option D is the SUM, not product. _Source: syllabus Mathematics, Ch 6 "Permutations and Combinations", §6.2 (Factorial Notation), p. 2–3._

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Find the number of arrangements of all the letters in the word 'CAT'.

  1. $3$
  2. $9$
  3. $27$
  4. $6$
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Correct answer: D. $6$

1. 'CAT' has 3 DISTINCT letters: C, A, T. 2. Number of ways to arrange $n$ distinct objects in a row: $n! = 3! = 6$. 3. Enumerate: CAT, CTA, ACT, ATC, TAC, TCA — six arrangements. 4. Option A is just $n$. Option C is $n^2$. Option D is $n^n$ (would apply to selection WITH repetition). _Source: syllabus Mathematics, Ch 6, §6.3 + Examples, p. 5–7._

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Find the number of distinct arrangements of all the letters in the word 'MISSISSIPPI'.

  1. $11!$
  2. $11!/(4!\,4!\,2!)$
  3. $\dfrac{11!}{2!}$
  4. $11^4$
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Correct answer: B. $11!/(4!\,4!\,2!)$

1. MISSISSIPPI has 11 letters: M(1), I(4), S(4), P(2). 2. Total letters: $11$. With ALL DISTINCT, the count would be $11!$. 3. But there are REPEATED letters. For each repeated letter group of size $k$, we OVERCOUNT by $k!$. 4. Adjust: divide by $4!$ (for the four I's), $4!$ (four S's), $2!$ (two P's), $1!$ (one M — trivial). 5. Result: $\dfrac{11!}{4!\,4!\,2!\,1!}$. Numerically: $39{,}916{,}800 / (24 \cdot 24 \cdot 2 \cdot 1) = 39{,}916{,}800/1152 = 34{,}650$. 6. Option A treats all letters as distinct (over-counts). Option C only adjusts for P's. Option D is unrelated. _Source: syllabus Mathematics, Ch 6, §6.3.2 (Permutations with repetition), p. 7._

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