A question paper is divided into 3 sections, A, B and C containing 3, 4 and 5 questions respectively. Find the number of ways of attempting six questions choosing at least one from each section.
Answer
Verified591.3k+ views
Hint: We can select one question each from two sections and four from the remaining section or two questions from each of the sections or two questions from one section, one question from another section and three questions from the remaining section
Hence form nine cases:
Case I: One question from section A, One from section B and four from section C
CaseII: One question from section A, four from section B and one from section C
Case III: Two questions from section A, Two from section B and Two from section C
Case IV: One question from section A, Two from section B and Three from section C
Case V: One question from section A, three from section B and two from section C
Case VI: Two questions from section A, three from section B and one from section C
Case VII: Two questions from section A, One from section B and three from section C
Case VIII: Three questions from section A, two from section B and one from section C
Case IX: Three questions from section A, One from section B and two from section C
Use the fundamental principle of counting, which states that if a task A can be done in m ways and task B can be done in n ways, then the number of ways of doing A and B is mn, and the number of ways of doing m or n is m+n. Hence determine the number of ways in which questions can be selected from each section.
Complete step-by-step answer:
Since at least one question from each section has to be selected the possible combinations of the number of questions in each section are permutations of (1,1,4),(2,2,2) and (1,2,3).
Hence, we have the following cases:
(Permutations of (1,1,4))
Case I: One question from section A, One from section B and four from section C
Hence, we select one question from section A, which can be done in $^{3}{{C}_{1}}=3$ ways, one from section B, which can be done in $^{4}{{C}_{1}}=4$ ways and 4 from section C which can be done in $^{5}{{C}_{4}}=5$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 4\times 5=60$
CaseII: One question from section A, four from section B and one from section C
Hence, we select one question from section A, which can be done in $^{3}{{C}_{1}}=3$ ways, four from section B, which can be done in $^{4}{{C}_{4}}=1$ ways and one from section C which can be done in $^{5}{{C}_{1}}=5$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 1\times 5=15$
(Permutations of (2,2,2))
Case III: Two questions from section A, Two from section B and Two from section C
Hence, we select two questions from section A, which can be done in $^{3}{{C}_{2}}=3$ ways, two from section B, which can be done in $^{4}{{C}_{2}}=6$ ways and two from section C which can be done in $^{5}{{C}_{2}}=10$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 6\times 10=180$
(Permutations of (1,2,3))
Case IV: One question from section A, Two from section B and Three from section C
Hence, we select one question from section A, which can be done in $^{3}{{C}_{1}}=3$ ways, two from section B, which can be done in $^{4}{{C}_{2}}=6$ ways and three from section C which can be done in $^{5}{{C}_{3}}=10$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 6\times 10=180$
Case V: One question from section A, three from section B and two from section C
Hence, we select one question from section A, which can be done in $^{3}{{C}_{1}}=3$ ways, three from section B, which can be done in $^{4}{{C}_{3}}=4$ ways and two from section C which can be done in $^{5}{{C}_{2}}=10$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 4\times 10=120$
Case VI: Two questions from section A, three from section B and one from section C
Hence, we select two questions from section A, which can be done in $^{3}{{C}_{2}}=3$ ways, three from section B, which can be done in $^{4}{{C}_{3}}=4$ ways and one from section C which can be done in $^{5}{{C}_{1}}=5$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 4\times 5=60$
Case VII: Two questions from section A, One from section B and three from section C
Hence, we select two questions from section A, which can be done in $^{3}{{C}_{2}}=3$ ways, one from section B, which can be done in $^{4}{{C}_{1}}=4$ ways and three from section C which can be done in $^{5}{{C}_{3}}=10$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 4\times 10=120$
Case VIII: Three questions from section A, two from section B and one from section C
Hence, we select three questions from section A, which can be done in $^{3}{{C}_{3}}=1$ way, two from section B, which can be done in $^{4}{{C}_{2}}=6$ ways and one from section C which can be done in $^{5}{{C}_{1}}=5$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=1\times 6\times 5=30$
Case IX: Three questions from section A, One from section B and two from section C
Hence, we select three questions from section A, which can be done in $^{3}{{C}_{3}}=1$ way, one from section B, which can be done in $^{4}{{C}_{1}}=4$ ways and two from section C which can be done in $^{5}{{C}_{2}}=10$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=1\times 4\times 10=40$
Hence, by the fundamental principle of counting the total number of ways in which six questions from sections A,B and C can be selected so that at least one question from each section is selected is $60+15+180+180+120+60+120+30+40=805$
Note: Alternative Solution: Best Method:
The number of ways in which 6 questions can be selected (without restriction) ${{=}^{12}}{{C}_{6}}=924$
Now, the number of cases in which no question from A is selected ${{=}^{9}}{{C}_{6}}=84$
The number of cases in which no question from B is selected ${{=}^{8}}{{C}_{6}}=28$
The number of cases in which no question from C is selected ${{=}^{7}}{{C}_{6}}=7$
Since there is no such case in which no question from two sections is selected, we have by the principle of inclusion and exclusion, the total number of ways of selecting six questions, such that at least one question from each section is included $=924-\left( 84+28+7 \right)=805$, which is the same as obtained above.
Hence form nine cases:
Case I: One question from section A, One from section B and four from section C
CaseII: One question from section A, four from section B and one from section C
Case III: Two questions from section A, Two from section B and Two from section C
Case IV: One question from section A, Two from section B and Three from section C
Case V: One question from section A, three from section B and two from section C
Case VI: Two questions from section A, three from section B and one from section C
Case VII: Two questions from section A, One from section B and three from section C
Case VIII: Three questions from section A, two from section B and one from section C
Case IX: Three questions from section A, One from section B and two from section C
Use the fundamental principle of counting, which states that if a task A can be done in m ways and task B can be done in n ways, then the number of ways of doing A and B is mn, and the number of ways of doing m or n is m+n. Hence determine the number of ways in which questions can be selected from each section.
Complete step-by-step answer:
Since at least one question from each section has to be selected the possible combinations of the number of questions in each section are permutations of (1,1,4),(2,2,2) and (1,2,3).
Hence, we have the following cases:
(Permutations of (1,1,4))
Case I: One question from section A, One from section B and four from section C
Hence, we select one question from section A, which can be done in $^{3}{{C}_{1}}=3$ ways, one from section B, which can be done in $^{4}{{C}_{1}}=4$ ways and 4 from section C which can be done in $^{5}{{C}_{4}}=5$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 4\times 5=60$
CaseII: One question from section A, four from section B and one from section C
Hence, we select one question from section A, which can be done in $^{3}{{C}_{1}}=3$ ways, four from section B, which can be done in $^{4}{{C}_{4}}=1$ ways and one from section C which can be done in $^{5}{{C}_{1}}=5$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 1\times 5=15$
(Permutations of (2,2,2))
Case III: Two questions from section A, Two from section B and Two from section C
Hence, we select two questions from section A, which can be done in $^{3}{{C}_{2}}=3$ ways, two from section B, which can be done in $^{4}{{C}_{2}}=6$ ways and two from section C which can be done in $^{5}{{C}_{2}}=10$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 6\times 10=180$
(Permutations of (1,2,3))
Case IV: One question from section A, Two from section B and Three from section C
Hence, we select one question from section A, which can be done in $^{3}{{C}_{1}}=3$ ways, two from section B, which can be done in $^{4}{{C}_{2}}=6$ ways and three from section C which can be done in $^{5}{{C}_{3}}=10$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 6\times 10=180$
Case V: One question from section A, three from section B and two from section C
Hence, we select one question from section A, which can be done in $^{3}{{C}_{1}}=3$ ways, three from section B, which can be done in $^{4}{{C}_{3}}=4$ ways and two from section C which can be done in $^{5}{{C}_{2}}=10$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 4\times 10=120$
Case VI: Two questions from section A, three from section B and one from section C
Hence, we select two questions from section A, which can be done in $^{3}{{C}_{2}}=3$ ways, three from section B, which can be done in $^{4}{{C}_{3}}=4$ ways and one from section C which can be done in $^{5}{{C}_{1}}=5$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 4\times 5=60$
Case VII: Two questions from section A, One from section B and three from section C
Hence, we select two questions from section A, which can be done in $^{3}{{C}_{2}}=3$ ways, one from section B, which can be done in $^{4}{{C}_{1}}=4$ ways and three from section C which can be done in $^{5}{{C}_{3}}=10$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=3\times 4\times 10=120$
Case VIII: Three questions from section A, two from section B and one from section C
Hence, we select three questions from section A, which can be done in $^{3}{{C}_{3}}=1$ way, two from section B, which can be done in $^{4}{{C}_{2}}=6$ ways and one from section C which can be done in $^{5}{{C}_{1}}=5$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=1\times 6\times 5=30$
Case IX: Three questions from section A, One from section B and two from section C
Hence, we select three questions from section A, which can be done in $^{3}{{C}_{3}}=1$ way, one from section B, which can be done in $^{4}{{C}_{1}}=4$ ways and two from section C which can be done in $^{5}{{C}_{2}}=10$ ways.
Hence by the fundamental principle of counting the total number of permutations in this case $=1\times 4\times 10=40$
Hence, by the fundamental principle of counting the total number of ways in which six questions from sections A,B and C can be selected so that at least one question from each section is selected is $60+15+180+180+120+60+120+30+40=805$
Note: Alternative Solution: Best Method:
The number of ways in which 6 questions can be selected (without restriction) ${{=}^{12}}{{C}_{6}}=924$
Now, the number of cases in which no question from A is selected ${{=}^{9}}{{C}_{6}}=84$
The number of cases in which no question from B is selected ${{=}^{8}}{{C}_{6}}=28$
The number of cases in which no question from C is selected ${{=}^{7}}{{C}_{6}}=7$
Since there is no such case in which no question from two sections is selected, we have by the principle of inclusion and exclusion, the total number of ways of selecting six questions, such that at least one question from each section is included $=924-\left( 84+28+7 \right)=805$, which is the same as obtained above.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success
Master Class 12 Maths: Engaging Questions & Answers for Success
Master Class 12 Biology: Engaging Questions & Answers for Success
Master Class 12 Physics: Engaging Questions & Answers for Success
Master Class 8 Maths: Engaging Questions & Answers for Success
Class 8 Question and Answer - Your Ultimate Solutions Guide
Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
What is the median of the first 10 natural numbers class 10 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Which of the following does not have a fundamental class 10 physics CBSE
State and prove converse of BPT Basic Proportionality class 10 maths CBSE