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Class 11MathematicsPermutations and CombinationsCombinationsMedium
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From (8) different pens and (7) different pencils (5) objects are to be selected with at least (1) pen and (1) pencil. How many ways are there?

Answer and explanation

Correct answer: (2982)

Total ways are (\binom{15}{5}=3003). Removing only pens (\binom{8}{5}=56) and only pencils (\binom{7}{5}=21) gives (2926).

Related tags

CombinationsClass11MediumPermutations And CombinationsMathematicsClass 11 Mcq
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10 more MCQs
  1. 1A shop has 6 different types of pencils and 2 different types of notebooks. In how many different ways can one pencil and one notebook be bought?
  2. 2A student chooses one subject from 3 subjects and one game from 5 games. How many different selections are possible in total?
  3. 3There are 4 bus routes and 3 train routes from a city to a village. In how many different ways can the journey be made using exactly one mode of transport?
  4. 4An examination contains 6 short-answer questions and 4 long-answer questions. A student has to solve exactly one of them. In how many ways can the student choose the question?
  5. 5A meal plate offers 2 bread choices, 3 vegetable choices, and 2 dessert choices. In how many ways can a plate be prepared by choosing one item from each category?
  6. 6A bag contains pens in 3 different colors, markers in 4 different colors, and pencils in 2 different colors. In how many ways can one pen, one marker, and one pencil be chosen?
  7. 7A vehicle number consists of two letters followed by two digits. There are 4 choices for each letter and 10 choices for each digit. If repetition of letters and digits is allowed, how many distinct vehicle numbers can be formed?
  8. 8A bag has 6 red and 5 blue balls. Four balls are to be selected with equal numbers of both colours. How many selections are possible?
  9. 9If \(\frac{(n+2)!}{(n-2)!}=3024\), what is the value of \(n\)?
  10. 10What is the value of \(\frac{10!}{2!\cdot 8!}\times\frac{8}{9}\)?
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