Total ways are \(\binom{19}{6}=27132\). Removing all-white \(\binom{9}{6}=84\) and all-black \(\binom{10}{6}=210\) gives (26838).
Step 2
Why this answer is correct
The correct answer is A. (26838). Total ways are \(\binom{19}{6}=27132\). Removing all-white \(\binom{9}{6}=84\) and all-black \(\binom{10}{6}=210\) gives (26838).
Step 3
Exam Tip
कुल \(\binom{19}{6}=27132\) हैं। सभी सफेद \(\binom{9}{6}=84\) और सभी काली \(\binom{10}{6}=210\) हटाने पर (26838) मिलते हैं।
The element (1) is fixed so choose (5) elements from (11). Subtracting \(\binom{9}{3}=84\) cases containing both (5), (6) gives (462-84=378).
Step 2
Why this answer is correct
The correct answer is B. (378). The element (1) is fixed so choose (5) elements from (11). Subtracting \(\binom{9}{3}=84\) cases containing both (5), (6) gives (462-84=378).
Step 3
Exam Tip
(1) तय है इसलिए (5) तत्व (11) में से चुनेंगे। (5), (6) दोनों होने पर \(\binom{9}{3}=84\) घटाने से (462-84=378) मिलता है।
The elements (1), (2) are fixed and (3), (4) are excluded. The remaining (4) elements are chosen from (9), so \(\binom{9}{4}=126\).
Step 2
Why this answer is correct
The correct answer is B. (126). The elements (1), (2) are fixed and (3), (4) are excluded. The remaining (4) elements are chosen from (9), so \(\binom{9}{4}=126\).
Step 3
Exam Tip
(1) और (2) तय हैं तथा (3), (4) हट गए हैं। बाकी (4) तत्व (9) में से चुने जाएंगे इसलिए \(\binom{9}{4}=126\) है।
By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{13}{6}\).
Step 2
Why this answer is correct
The correct answer is A. \(\binom{13}{6}\). By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{13}{6}\).
Step 3
Exam Tip
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{13}{6}\) है।
The number of pens can be (2), (3), (4), or (5). The total is \(\binom{10}{2}\binom{9}{5}+\binom{10}{3}\binom{9}{4}+\binom{10}{4}\binom{9}{3}+\binom{10}{5}\binom{9}{2}=47502\).
Step 2
Why this answer is correct
The correct answer is C. (47502). The number of pens can be (2), (3), (4), or (5). The total is \(\binom{10}{2}\binom{9}{5}+\binom{10}{3}\binom{9}{4}+\binom{10}{4}\binom{9}{3}+\binom{10}{5}\binom{9}{2}=47502\).
Step 3
Exam Tip
पेन (2), (3), (4) या (5) हो सकते हैं। कुल \(\binom{10}{2}\binom{9}{5}+\binom{10}{3}\binom{9}{4}+\binom{10}{4}\binom{9}{3}+\binom{10}{5}\binom{9}{2}=47502\) है।
Total ways are \(\binom{16}{7}=11440\). Removing selections with (0) and (1) special student gives \(11440-\binom{11}{7}-\binom{5}{1}\binom{11}{6}=8800\).
Step 2
Why this answer is correct
The correct answer is C. (8800). Total ways are \(\binom{16}{7}=11440\). Removing selections with (0) and (1) special student gives \(11440-\binom{11}{7}-\binom{5}{1}\binom{11}{6}=8800\).
Step 3
Exam Tip
कुल \(\binom{16}{7}=11440\) हैं। (0) और (1) विशेष वाले चयन हटाने पर \(11440-\binom{11}{7}-\binom{5}{1}\binom{11}{6}=8800\) है।
One color is fixed and one is excluded. Choose the remaining (5) colors from (10) and subtract \(\binom{8}{5}\) selections missing both other colors to get (196).
Step 2
Why this answer is correct
The correct answer is B. (196). One color is fixed and one is excluded. Choose the remaining (5) colors from (10) and subtract \(\binom{8}{5}\) selections missing both other colors to get (196).
Step 3
Exam Tip
एक रंग तय और एक हट गया है। बाकी (5) रंग (10) में से चुनें और दोनों अन्य रंग न आने वाले \(\binom{8}{5}\) चयन घटाएं तो (196) मिलेगा।
The number of doctors will be (4), (5), or (6). The total is \(\binom{10}{4}\binom{8}{2}+\binom{10}{5}\binom{8}{1}+\binom{10}{6}=8106\).
Step 2
Why this answer is correct
The correct answer is B. (8106). The number of doctors will be (4), (5), or (6). The total is \(\binom{10}{4}\binom{8}{2}+\binom{10}{5}\binom{8}{1}+\binom{10}{6}=8106\).
Step 3
Exam Tip
डॉक्टरों की संख्या (4), (5) या (6) होगी। कुल \(\binom{10}{4}\binom{8}{2}+\binom{10}{5}\binom{8}{1}+\binom{10}{6}=8106\) है।
The number of English books can be (2), (3), or (4). The total is \(\binom{9}{2}\binom{10}{4}+\binom{9}{3}\binom{10}{3}+\binom{9}{4}\binom{10}{2}=23310\).
Step 2
Why this answer is correct
The correct answer is C. (23310). The number of English books can be (2), (3), or (4). The total is \(\binom{9}{2}\binom{10}{4}+\binom{9}{3}\binom{10}{3}+\binom{9}{4}\binom{10}{2}=23310\).
Step 3
Exam Tip
अंग्रेजी पुस्तकों की संख्या (2), (3) या (4) हो सकती है। कुल \(\binom{9}{2}\binom{10}{4}+\binom{9}{3}\binom{10}{3}+\binom{9}{4}\binom{10}{2}=23310\) है।
The cases are (4), (5), and (6) mathematics books. The total is \(\binom{11}{4}\binom{8}{2}+\binom{11}{5}\binom{8}{1}+\binom{11}{6}=13398\).
Step 2
Why this answer is correct
The correct answer is C. (13398). The cases are (4), (5), and (6) mathematics books. The total is \(\binom{11}{4}\binom{8}{2}+\binom{11}{5}\binom{8}{1}+\binom{11}{6}=13398\).
Step 3
Exam Tip
मामले (4), (5) और (6) गणित पुस्तकों के हैं। कुल \(\binom{11}{4}\binom{8}{2}+\binom{11}{5}\binom{8}{1}+\binom{11}{6}=13398\) है।
Total ways are \(\binom{12}{6}=924\) and ways with both special persons are \(\binom{10}{4}=210\). Hence (924-210=714).
Step 2
Why this answer is correct
The correct answer is A. (714). Total ways are \(\binom{12}{6}=924\) and ways with both special persons are \(\binom{10}{4}=210\). Hence (924-210=714).
Step 3
Exam Tip
कुल \(\binom{12}{6}=924\) हैं और दोनों विशेष साथ हों तो \(\binom{10}{4}=210\) हैं। इसलिए (924-210=714) तरीके हैं।
Total pairs are \(\binom{20}{2}=190\). Replacing \(\binom{8}{2}\) and \(\binom{5}{2}\) by (1), (1) for collinear groups gives (154).
Step 2
Why this answer is correct
The correct answer is B. (154). Total pairs are \(\binom{20}{2}=190\). Replacing \(\binom{8}{2}\) and \(\binom{5}{2}\) by (1), (1) for collinear groups gives (154).
Step 3
Exam Tip
कुल \(\binom{20}{2}=190\) जोड़ियां हैं। समरेखीय समूहों के लिए \(\binom{8}{2}\) और \(\binom{5}{2}\) के स्थान पर (1), (1) रेखा लेने से (154) मिलता है।
Total ways are \(\binom{14}{6}=3003\) and ways with both special books are \(\binom{12}{4}=495\). Hence (3003-495=2508).
Step 2
Why this answer is correct
The correct answer is A. (2508). Total ways are \(\binom{14}{6}=3003\) and ways with both special books are \(\binom{12}{4}=495\). Hence (3003-495=2508).
Step 3
Exam Tip
कुल \(\binom{14}{6}=3003\) हैं और दोनों विशेष साथ हों तो \(\binom{12}{4}=495\) हैं। इसलिए (3003-495=2508) तरीके हैं।
You can choose (3), (4), or (5) from the first (5) questions. The total is \(\binom{5}{3}\binom{7}{4}+\binom{5}{4}\binom{7}{3}+\binom{5}{5}\binom{7}{2}=462\).
Step 2
Why this answer is correct
The correct answer is C. (462). You can choose (3), (4), or (5) from the first (5) questions. The total is \(\binom{5}{3}\binom{7}{4}+\binom{5}{4}\binom{7}{3}+\binom{5}{5}\binom{7}{2}=462\).
Step 3
Exam Tip
पहले (5) में से (3), (4) या (5) प्रश्न चुने जा सकते हैं। कुल \(\binom{5}{3}\binom{7}{4}+\binom{5}{4}\binom{7}{3}+\binom{5}{5}\binom{7}{2}=462\) है।
