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Permutations and Combinations question

2003 · Shift 0 · Q105
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Permutations and Combinations question

2003 · Shift 0 · Q105

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
  1. A
    346
  2. B
    140
  3. C
    196
  4. D
    280
View written solutionFree

Correct answer: C

Let the first 5 questions be Group I, and the remaining 8 questions be Group II.

The student must answer 10 out of 13 questions, with the condition that he chooses at least 4 from the first 5.

We count the valid selections case-wise.

1. Possible numbers chosen from the first 5

Since at least 4 must be chosen from the first 5, the possibilities are:

  • choose 4 from the first 5
  • choose 5 from the first 5

Because total questions answered = 10, the remaining must come from the other 8.

2. Case 1: Choose 4 from the first 5

Number of ways:

(54)(86)\binom{5}{4}\binom{8}{6}(45​)(68​)

Since after choosing 4 from first 5, we need 6 more from the remaining 8.

Compute:

(54)=5,(86)=(82)=28\binom{5}{4}=5, \qquad \binom{8}{6}=\binom{8}{2}=28(45​)=5,(68​)=(28​)=28

So,

5×28=1405 \times 28 = 1405×28=140

3. Case 2: Choose 5 from the first 5

Number of ways:

(55)(85)\binom{5}{5}\binom{8}{5}(55​)(58​)

Since after choosing all 5 from first 5, we need 5 more from the remaining 8.

Compute:

(55)=1,(85)=(83)=56\binom{5}{5}=1, \qquad \binom{8}{5}=\binom{8}{3}=56(55​)=1,(58​)=(38​)=56

So,

1×56=561 \times 56 = 561×56=56

4. Total number of choices

140+56=196140 + 56 = 196140+56=196

So the required number of choices is:

196\boxed{196}196​

Thus, Option C is correct.

5. Comparison with stored correct answer

Stored correct answer: C

Our derived answer is also C (196), so they agree.

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