The number of women can be (3), (4), (5), or (6). The correct sum is \(\binom{8}{3}\binom{9}{3}+\binom{8}{4}\binom{9}{2}+\binom{8}{5}\binom{9}{1}+\binom{8}{6}=7756\).
Step 2
Why this answer is correct
The correct answer is D. (7756). The number of women can be (3), (4), (5), or (6). The correct sum is \(\binom{8}{3}\binom{9}{3}+\binom{8}{4}\binom{9}{2}+\binom{8}{5}\binom{9}{1}+\binom{8}{6}=7756\).
Step 3
Exam Tip
महिलाएं (3), (4), (5) या (6) हो सकती हैं। सही योग \(\binom{8}{3}\binom{9}{3}+\binom{8}{4}\binom{9}{2}+\binom{8}{5}\binom{9}{1}+\binom{8}{6}=7756\) है।
You can choose (3), (4), or (5) from the first (5) questions. The total is \(\binom{5}{3}\binom{8}{5}+\binom{5}{4}\binom{8}{4}+\binom{5}{5}\binom{8}{3}=966\).
Step 2
Why this answer is correct
The correct answer is C. (966). You can choose (3), (4), or (5) from the first (5) questions. The total is \(\binom{5}{3}\binom{8}{5}+\binom{5}{4}\binom{8}{4}+\binom{5}{5}\binom{8}{3}=966\).
Step 3
Exam Tip
पहले (5) में से (3), (4) या (5) प्रश्न चुने जा सकते हैं। कुल \(\binom{5}{3}\binom{8}{5}+\binom{5}{4}\binom{8}{4}+\binom{5}{5}\binom{8}{3}=966\) है।
Total ways are \(\binom{15}{6}=5005\) and ways with both special books are \(\binom{13}{4}=715\). Hence (5005-715=4290).
Step 2
Why this answer is correct
The correct answer is A. (4290). Total ways are \(\binom{15}{6}=5005\) and ways with both special books are \(\binom{13}{4}=715\). Hence (5005-715=4290).
Step 3
Exam Tip
कुल \(\binom{15}{6}=5005\) हैं और दोनों विशेष साथ हों तो \(\binom{13}{4}=715\) हैं। इसलिए (5005-715=4290) तरीके हैं।
Total pairs are \(\binom{22}{2}=231\). Replacing \(\binom{9}{2}\) and \(\binom{6}{2}\) by (1), (1) for collinear groups gives (182).
Step 2
Why this answer is correct
The correct answer is B. (182). Total pairs are \(\binom{22}{2}=231\). Replacing \(\binom{9}{2}\) and \(\binom{6}{2}\) by (1), (1) for collinear groups gives (182).
Step 3
Exam Tip
कुल \(\binom{22}{2}=231\) जोड़ियां हैं। समरेखीय समूहों में \(\binom{9}{2}\) और \(\binom{6}{2}\) की जगह (1), (1) रेखा लेने से (182) मिलता है।
Total ways are \(\binom{13}{6}=1716\) and ways with both special persons are \(\binom{11}{4}=330\). Hence (1716-330=1386).
Step 2
Why this answer is correct
The correct answer is A. (1386). Total ways are \(\binom{13}{6}=1716\) and ways with both special persons are \(\binom{11}{4}=330\). Hence (1716-330=1386).
Step 3
Exam Tip
कुल \(\binom{13}{6}=1716\) हैं और दोनों विशेष साथ हों तो \(\binom{11}{4}=330\) हैं। इसलिए (1716-330=1386) तरीके हैं।
The number of doctors will be (4), (5), (6), or (7). The total is \(\binom{11}{4}\binom{8}{3}+\binom{11}{5}\binom{8}{2}+\binom{11}{6}\binom{8}{1}+\binom{11}{7}=35442\).
Step 2
Why this answer is correct
The correct answer is D. (35442). The number of doctors will be (4), (5), (6), or (7). The total is \(\binom{11}{4}\binom{8}{3}+\binom{11}{5}\binom{8}{2}+\binom{11}{6}\binom{8}{1}+\binom{11}{7}=35442\).
Step 3
Exam Tip
डॉक्टरों की संख्या (4), (5), (6) या (7) होगी। कुल \(\binom{11}{4}\binom{8}{3}+\binom{11}{5}\binom{8}{2}+\binom{11}{6}\binom{8}{1}+\binom{11}{7}=35442\) है।
One color is fixed and one is excluded. Choose the remaining (5) colors from (11) and subtract \(\binom{9}{5}\) selections missing both other colors to get (336).
Step 2
Why this answer is correct
The correct answer is A. (336). One color is fixed and one is excluded. Choose the remaining (5) colors from (11) and subtract \(\binom{9}{5}\) selections missing both other colors to get (336).
Step 3
Exam Tip
एक रंग तय और एक हट गया है। बाकी (5) रंग (11) में से चुनें और दोनों अन्य रंग न आने वाले \(\binom{9}{5}\) चयन घटाएं तो (336) मिलेगा।
The number of special students can be (3), (4), (5), or (6). The total is \(\binom{6}{3}\binom{11}{5}+\binom{6}{4}\binom{11}{4}+\binom{6}{5}\binom{11}{3}+\binom{6}{6}\binom{11}{2}=15235\).
Step 2
Why this answer is correct
The correct answer is D. (15235). The number of special students can be (3), (4), (5), or (6). The total is \(\binom{6}{3}\binom{11}{5}+\binom{6}{4}\binom{11}{4}+\binom{6}{5}\binom{11}{3}+\binom{6}{6}\binom{11}{2}=15235\).
Step 3
Exam Tip
विशेष छात्र (3), (4), (5) या (6) हो सकते हैं। कुल \(\binom{6}{3}\binom{11}{5}+\binom{6}{4}\binom{11}{4}+\binom{6}{5}\binom{11}{3}+\binom{6}{6}\binom{11}{2}=15235\) है।
The number of pens can be (2), (3), (4), or (5). The total is \(\binom{10}{2}\binom{8}{5}+\binom{10}{3}\binom{8}{4}+\binom{10}{4}\binom{8}{3}+\binom{10}{5}\binom{8}{2}=29736\).
Step 2
Why this answer is correct
The correct answer is B. (29736). The number of pens can be (2), (3), (4), or (5). The total is \(\binom{10}{2}\binom{8}{5}+\binom{10}{3}\binom{8}{4}+\binom{10}{4}\binom{8}{3}+\binom{10}{5}\binom{8}{2}=29736\).
Step 3
Exam Tip
पेन (2), (3), (4) या (5) हो सकते हैं। कुल \(\binom{10}{2}\binom{8}{5}+\binom{10}{3}\binom{8}{4}+\binom{10}{4}\binom{8}{3}+\binom{10}{5}\binom{8}{2}=29736\) है।
By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{14}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(\binom{14}{7}\). By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{14}{7}\).
Step 3
Exam Tip
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{14}{7}\) है।
The elements (1), (2) are fixed and (3), (4) are excluded. The remaining (5) elements are chosen from (10), so \(\binom{10}{5}=252\).
Step 2
Why this answer is correct
The correct answer is B. (252). The elements (1), (2) are fixed and (3), (4) are excluded. The remaining (5) elements are chosen from (10), so \(\binom{10}{5}=252\).
Step 3
Exam Tip
(1) और (2) तय हैं तथा (3), (4) हट गए हैं। बाकी (5) तत्व (10) में से चुने जाएंगे इसलिए \(\binom{10}{5}=252\) है।
The element (1) is fixed so choose (5) elements from (12). Subtracting \(\binom{10}{3}=120\) cases containing both (5), (6) gives (792-120=672).
Step 2
Why this answer is correct
The correct answer is A. (672). The element (1) is fixed so choose (5) elements from (12). Subtracting \(\binom{10}{3}=120\) cases containing both (5), (6) gives (792-120=672).
Step 3
Exam Tip
(1) तय है इसलिए (5) तत्व (12) में से चुनेंगे। (5), (6) दोनों होने पर \(\binom{10}{3}=120\) घटाने से (792-120=672) मिलता है।
Total ways are \(\binom{21}{7}=116280\). Removing all-white \(\binom{10}{7}=120\) and all-black \(\binom{11}{7}=330\) gives (115830).
Step 2
Why this answer is correct
The correct answer is B. (115830). Total ways are \(\binom{21}{7}=116280\). Removing all-white \(\binom{10}{7}=120\) and all-black \(\binom{11}{7}=330\) gives (115830).
Step 3
Exam Tip
कुल \(\binom{21}{7}=116280\) हैं। सभी सफेद \(\binom{10}{7}=120\) और सभी काली \(\binom{11}{7}=330\) हटाने पर (115830) मिलते हैं।
The letter (a) is fixed and (b) is excluded. Choose (1) from (c,d) and (4) from the remaining (9), so \(\binom{2}{1}\binom{9}{4}=252\).
Step 2
Why this answer is correct
The correct answer is C. (252). The letter (a) is fixed and (b) is excluded. Choose (1) from (c,d) and (4) from the remaining (9), so \(\binom{2}{1}\binom{9}{4}=252\).
Step 3
Exam Tip
(a) तय है और (b) हट गया है। (c,d) में से (1) और शेष (9) में से (4) चुनेंगे इसलिए \(\binom{2}{1}\binom{9}{4}=252\) है।
Chemistry is fixed and biology is excluded. Choose the remaining (4) subjects from (9) and subtract \(\binom{7}{2}\) selections containing both mathematics and physics to get (105).
Step 2
Why this answer is correct
The correct answer is A. (105). Chemistry is fixed and biology is excluded. Choose the remaining (4) subjects from (9) and subtract \(\binom{7}{2}\) selections containing both mathematics and physics to get (105).
Step 3
Exam Tip
रसायन तय और जीवविज्ञान हट गया है। बाकी (4) विषय (9) में से चुनें और गणित-भौतिकी दोनों वाले \(\binom{7}{2}\) चयन घटाएं तो (105) मिलते हैं।
The (3) numbers are fixed and (2) are excluded. The remaining (5) numbers are chosen from (13), so \(\binom{13}{5}=1287\).
Step 2
Why this answer is correct
The correct answer is B. (1287). The (3) numbers are fixed and (2) are excluded. The remaining (5) numbers are chosen from (13), so \(\binom{13}{5}=1287\).
Step 3
Exam Tip
(3) संख्याएं तय हैं और (2) हट गई हैं। बाकी (5) संख्याएं (13) में से चुनी जाएंगी इसलिए \(\binom{13}{5}=1287\) है।
For a failed triangle (3) points are chosen from the (10) collinear points. So the number is \(\binom{10}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(\binom{10}{3}\). For a failed triangle (3) points are chosen from the (10) collinear points. So the number is \(\binom{10}{3}\).
Step 3
Exam Tip
असफल त्रिभुज के लिए (10) समरेखीय बिंदुओं में से (3) चुने जाते हैं। इसलिए संख्या \(\binom{10}{3}\) है।
If all special students are included there are \(\binom{11}{4}=330\) ways and if all are excluded there are \(\binom{11}{7}=330\) ways. The total is (660).
Step 2
Why this answer is correct
The correct answer is B. (660). If all special students are included there are \(\binom{11}{4}=330\) ways and if all are excluded there are \(\binom{11}{7}=330\) ways. The total is (660).
Step 3
Exam Tip
सभी विशेष शामिल हों तो \(\binom{11}{4}=330\) और सभी बाहर हों तो \(\binom{11}{7}=330\) तरीके हैं। कुल (660) है।
The number of seniors can be (0), (1), (2), or (3). The total is \(\binom{10}{0}\binom{12}{7}+\binom{10}{1}\binom{12}{6}+\binom{10}{2}\binom{12}{5}+\binom{10}{3}\binom{12}{4}=105072\).
Step 2
Why this answer is correct
The correct answer is D. (105072). The number of seniors can be (0), (1), (2), or (3). The total is \(\binom{10}{0}\binom{12}{7}+\binom{10}{1}\binom{12}{6}+\binom{10}{2}\binom{12}{5}+\binom{10}{3}\binom{12}{4}=105072\).
Step 3
Exam Tip
वरिष्ठों की संख्या (0), (1), (2) या (3) हो सकती है। कुल \(\binom{10}{0}\binom{12}{7}+\binom{10}{1}\binom{12}{6}+\binom{10}{2}\binom{12}{5}+\binom{10}{3}\binom{12}{4}=105072\) है।
Choose (1) of the two fixed cards and (5) cards from the remaining (13). The ways are \(\binom{2}{1}\binom{13}{5}=2574\).
Step 2
Why this answer is correct
The correct answer is A. (2574). Choose (1) of the two fixed cards and (5) cards from the remaining (13). The ways are \(\binom{2}{1}\binom{13}{5}=2574\).
Step 3
Exam Tip
दो निश्चित कार्डों में से (1) चुनें और बाकी (5) कार्ड (13) में से चुनें। तरीके \(\binom{2}{1}\binom{13}{5}=2574\) हैं।
The number of odd selections is \(2^{15-1}=16384\). Remember that even and odd selections are equal in such questions.
Step 2
Why this answer is correct
The correct answer is C. (16384). The number of odd selections is \(2^{15-1}=16384\). Remember that even and odd selections are equal in such questions.
Step 3
Exam Tip
विषम चयन की संख्या \(2^{15-1}=16384\) होती है। परीक्षा में सम और विषम चयन की संख्या बराबर याद रखें।
(14) खिलाड़ियों में से (7) खिलाड़ियों का चयन करना है। एक कप्तान पहले से तय है और उसे चुना नहीं जाना है लेकिन उपकप्तान अवश्य चुना जाना है। कितने तरीके हैं?
The captain is excluded and the vice-captain is fixed. The remaining (6) players are chosen from (12), so \(\binom{12}{6}=924\).
Step 2
Why this answer is correct
The correct answer is D. (924). The captain is excluded and the vice-captain is fixed. The remaining (6) players are chosen from (12), so \(\binom{12}{6}=924\).
Step 3
Exam Tip
कप्तान हट गया और उपकप्तान तय है। बाकी (6) खिलाड़ी (12) में से चुने जाएंगे इसलिए \(\binom{12}{6}=924\) है।
The number of special objects can be (0), (1), or (2). The total is \(\binom{7}{0}\binom{13}{8}+\binom{7}{1}\binom{13}{7}+\binom{7}{2}\binom{13}{6}=49335\).
Step 2
Why this answer is correct
The correct answer is C. (49335). The number of special objects can be (0), (1), or (2). The total is \(\binom{7}{0}\binom{13}{8}+\binom{7}{1}\binom{13}{7}+\binom{7}{2}\binom{13}{6}=49335\).
Step 3
Exam Tip
विशेष वस्तुएं (0), (1) या (2) चुनी जा सकती हैं। कुल \(\binom{7}{0}\binom{13}{8}+\binom{7}{1}\binom{13}{7}+\binom{7}{2}\binom{13}{6}=49335\) है।
