The cases are (3), (4), and (5) chemistry books. The total is \(\binom{9}{3}\binom{6}{2}+\binom{9}{4}\binom{6}{1}+\binom{9}{5}=1932\).
Step 2
Why this answer is correct
The correct answer is C. (1932). The cases are (3), (4), and (5) chemistry books. The total is \(\binom{9}{3}\binom{6}{2}+\binom{9}{4}\binom{6}{1}+\binom{9}{5}=1932\).
Step 3
Exam Tip
मामले (3), (4) और (5) रसायन पुस्तकों के हैं। कुल \(\binom{9}{3}\binom{6}{2}+\binom{9}{4}\binom{6}{1}+\binom{9}{5}=1932\) है।
Total triples are \(\binom{14}{3}=364\) and \(\binom{6}{3}=20\) collinear triples do not form triangles. Hence (364-20=344).
Step 2
Why this answer is correct
The correct answer is A. (344). Total triples are \(\binom{14}{3}=364\) and \(\binom{6}{3}=20\) collinear triples do not form triangles. Hence (364-20=344).
Step 3
Exam Tip
कुल \(\binom{14}{3}=364\) त्रिक हैं और \(\binom{6}{3}=20\) समरेखीय त्रिक त्रिभुज नहीं बनाते। इसलिए (364-20=344) है।
Total pairs are \(\binom{18}{2}=153\) and (7) collinear points give (1) line instead of \(\binom{7}{2}\). Hence (153-21+1=133).
Step 2
Why this answer is correct
The correct answer is B. (133). Total pairs are \(\binom{18}{2}=153\) and (7) collinear points give (1) line instead of \(\binom{7}{2}\). Hence (153-21+1=133).
Step 3
Exam Tip
कुल \(\binom{18}{2}=153\) जोड़ियां हैं और (7) समरेखीय बिंदु \(\binom{7}{2}\) के स्थान पर (1) रेखा देते हैं। इसलिए (153-21+1=133) है।
Total ways are \(\binom{13}{5}=1287\) and ways with both special books are \(\binom{11}{3}=165\). Hence (1287-165=1122).
Step 2
Why this answer is correct
The correct answer is B. (1188). Total ways are \(\binom{13}{5}=1287\) and ways with both special books are \(\binom{11}{3}=165\). Hence (1287-165=1122).
Step 3
Exam Tip
कुल \(\binom{13}{5}=1287\) हैं और दोनों विशेष साथ हों तो \(\binom{11}{3}=165\) हैं। इसलिए (1287-165=1122) तरीके हैं।
The cases are choosing (2), (3), or (4) from the first (4). The total is \(\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2}=371\).
Step 2
Why this answer is correct
The correct answer is C. (371). The cases are choosing (2), (3), or (4) from the first (4). The total is \(\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2}=371\).
Step 3
Exam Tip
मामले पहले (4) में से (2), (3) या (4) चुनने के हैं। कुल \(\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2}=371\) है।
The cases are (2), (3), (4), and (5) women. The total is \(\binom{8}{2}\binom{8}{3}+\binom{8}{3}\binom{8}{2}+\binom{8}{4}\binom{8}{1}+\binom{8}{5}=3752\).
Step 2
Why this answer is correct
The correct answer is D. (3752). The cases are (2), (3), (4), and (5) women. The total is \(\binom{8}{2}\binom{8}{3}+\binom{8}{3}\binom{8}{2}+\binom{8}{4}\binom{8}{1}+\binom{8}{5}=3752\).
Step 3
Exam Tip
मामले (2), (3), (4) और (5) महिलाओं के हैं। कुल \(\binom{8}{2}\binom{8}{3}+\binom{8}{3}\binom{8}{2}+\binom{8}{4}\binom{8}{1}+\binom{8}{5}=3752\) है।
Either (0) or (1) special object can be chosen. The total is \(\binom{4}{0}\binom{9}{5}+\binom{4}{1}\binom{9}{4}=630\).
Step 2
Why this answer is correct
The correct answer is C. (882). Either (0) or (1) special object can be chosen. The total is \(\binom{4}{0}\binom{9}{5}+\binom{4}{1}\binom{9}{4}=630\).
Step 3
Exam Tip
विशेष वस्तुओं में से (0) या (1) चुनी जा सकती है। कुल \(\binom{4}{0}\binom{9}{5}+\binom{4}{1}\binom{9}{4}=630\) है।
The number of seniors can be (0), (1), or (2). The total is \(\binom{6}{0}\binom{8}{5}+\binom{6}{1}\binom{8}{4}+\binom{6}{2}\binom{8}{3}=1316\).
Step 2
Why this answer is correct
The correct answer is C. (896). The number of seniors can be (0), (1), or (2). The total is \(\binom{6}{0}\binom{8}{5}+\binom{6}{1}\binom{8}{4}+\binom{6}{2}\binom{8}{3}=1316\).
Step 3
Exam Tip
वरिष्ठों की संख्या (0), (1) या (2) हो सकती है। कुल \(\binom{6}{0}\binom{8}{5}+\binom{6}{1}\binom{8}{4}+\binom{6}{2}\binom{8}{3}=1316\) है।
The cases are choosing (2), (3), or (4) from the last (4). The total is \(\binom{4}{2}\binom{5}{3}+\binom{4}{3}\binom{5}{2}+\binom{4}{4}\binom{5}{1}=105\).
Step 2
Why this answer is correct
The correct answer is C. (101). The cases are choosing (2), (3), or (4) from the last (4). The total is \(\binom{4}{2}\binom{5}{3}+\binom{4}{3}\binom{5}{2}+\binom{4}{4}\binom{5}{1}=105\).
Step 3
Exam Tip
मामले अंतिम (4) में से (2), (3) या (4) चुनने के हैं। कुल \(\binom{4}{2}\binom{5}{3}+\binom{4}{3}\binom{5}{2}+\binom{4}{4}\binom{5}{1}=105\) है।
If both are included there are \(\binom{8}{3}=56\) ways and if both are excluded there are \(\binom{8}{5}=56\) ways. The total is (112).
Step 2
Why this answer is correct
The correct answer is B. (126). If both are included there are \(\binom{8}{3}=56\) ways and if both are excluded there are \(\binom{8}{5}=56\) ways. The total is (112).
Step 3
Exam Tip
दोनों शामिल हों तो \(\binom{8}{3}=56\) और दोनों बाहर हों तो \(\binom{8}{5}=56\) तरीके हैं। कुल (56+56=112) है।
For a failed triangle (3) points are chosen from the (6) collinear points. So the number is \(\binom{6}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\binom{6}{3}\). For a failed triangle (3) points are chosen from the (6) collinear points. So the number is \(\binom{6}{3}\).
Step 3
Exam Tip
असफल त्रिभुज के लिए (6) समरेखीय बिंदुओं में से (3) चुने जाते हैं। इसलिए संख्या \(\binom{6}{3}\) है।
The (2) numbers are fixed and (1) is excluded. The remaining (3) numbers are chosen from (9) in \(\binom{9}{3}=84\) ways.
Step 2
Why this answer is correct
The correct answer is A. (84). The (2) numbers are fixed and (1) is excluded. The remaining (3) numbers are chosen from (9) in \(\binom{9}{3}=84\) ways.
Step 3
Exam Tip
(2) संख्याएं तय हैं और (1) हट गई है। बाकी (3) संख्याएं (9) में से \(\binom{9}{3}=84\) तरीकों से चुनी जाएंगी।
Total ways are \(\binom{7}{4}=35\) and if both special subjects are included there are \(\binom{5}{2}=10\) ways. Hence (35-10=25).
Step 2
Why this answer is correct
The correct answer is C. (25). Total ways are \(\binom{7}{4}=35\) and if both special subjects are included there are \(\binom{5}{2}=10\) ways. Hence (35-10=25).
Step 3
Exam Tip
कुल \(\binom{7}{4}=35\) हैं और दोनों विशेष विषय साथ हों तो \(\binom{5}{2}=10\) हैं। इसलिए (35-10=25) है।
The letter (a) is fixed and (b) is excluded so choose the remaining (3) letters from (6). The ways are \(\binom{6}{3}=20\).
Step 2
Why this answer is correct
The correct answer is C. (20). The letter (a) is fixed and (b) is excluded so choose the remaining (3) letters from (6). The ways are \(\binom{6}{3}=20\).
Step 3
Exam Tip
(a) तय है और (b) हट गया है इसलिए बाकी (3) अक्षर (6) में से चुनेंगे। तरीके \(\binom{6}{3}=20\) हैं।
