Adding the remaining factor ((n-r)!) makes the numerator (n!). In exams identify the missing tail while forming factorial form.
Step 2
Why this answer is correct
The correct answer is B. ((n-r)!). Adding the remaining factor ((n-r)!) makes the numerator (n!). In exams identify the missing tail while forming factorial form.
Step 3
Exam Tip
बाकी गुणनखंड ((n-r)!) जोड़ने से ऊपर (n!) बनता है। परीक्षा में factorial रूप बनाते समय missing tail पहचानें।
A. चुनी गई (r) वस्तुओं के अंदरूनी क्रम/Internal orders of selected (r) objects
Step 1
Concept
In combination the order of selected objects is not considered different. In exams remove internal order in unordered selection.
Step 2
Why this answer is correct
The correct answer is A. चुनी गई (r) वस्तुओं के अंदरूनी क्रम / Internal orders of selected (r) objects. In combination the order of selected objects is not considered different. In exams remove internal order in unordered selection.
Step 3
Exam Tip
Combination में चुनी वस्तुओं का क्रम अलग नहीं माना जाता। परीक्षा में unordered selection में internal order हटाएँ।
First there are \(^{n}C_r\) selections and then (r!) arrangements for each selection. In exams multiply for choose then arrange.
Step 2
Why this answer is correct
The correct answer is C. \(^{n}C_r\times r!\). First there are \(^{n}C_r\) selections and then (r!) arrangements for each selection. In exams multiply for choose then arrange.
Step 3
Exam Tip
पहले चयन \(^{n}C_r\) और फिर हर चयन की (r!) व्यवस्थाएँ होती हैं। परीक्षा में choose then arrange को multiply करें।
For four ordered positions the choices are (9), (8), (7), (6). In exams each next choice decreases by one without repetition.
Step 2
Why this answer is correct
The correct answer is A. \(9\times8\times7\times6\). For four ordered positions the choices are (9), (8), (7), (6). In exams each next choice decreases by one without repetition.
Step 3
Exam Tip
चार क्रमबद्ध स्थानों के लिए विकल्प (9), (8), (7), (6) हैं। परीक्षा में बिना पुनरावृत्ति हर अगला विकल्प एक कम होता है।
B. हर unordered pair दो orders में गिना गया है/Each unordered pair is counted in two orders
Step 1
Concept
(AB) and (BA) are the same unordered pair. In exams divide by (2!) when there is no direction.
Step 2
Why this answer is correct
The correct answer is B. हर unordered pair दो orders में गिना गया है / Each unordered pair is counted in two orders. (AB) and (BA) are the same unordered pair. In exams divide by (2!) when there is no direction.
Step 3
Exam Tip
(AB) और (BA) एक ही unordered pair हैं। परीक्षा में direction न हो तो (2!) से divide करें।
Simplifying the factorial form gives the factor \(\frac{n}{r}\). In exams check adjacent upper index identities by ratio.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}C_r=\frac{n}{r},^{n-1}C_{r-1}\). Simplifying the factorial form gives the factor \(\frac{n}{r}\). In exams check adjacent upper index identities by ratio.
Step 3
Exam Tip
Factorial form से सरल करने पर \(\frac{n}{r}\) factor आता है। परीक्षा में adjacent upper index identities को ratio से जाँचें।
Put (n=8) and (r=4) in \(^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}\). In exams the upper index of both terms decreases by (1).
Step 2
Why this answer is correct
The correct answer is A. \(^{7}C_4+^{7}C_3\). Put (n=8) and (r=4) in \(^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}\). In exams the upper index of both terms decreases by (1).
Step 3
Exam Tip
\(^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}\) में (n=8) और (r=4) रखें। परीक्षा में दोनों terms का upper index (1) कम होता है।
A. (b) को (r) brackets से चुनना/Choosing (b) from (r) brackets
Step 1
Concept
There are \(^{n}C_r\) ways to choose (b) from (r) of the (n) brackets. In exams treat a binomial coefficient as selection.
Step 2
Why this answer is correct
The correct answer is A. (b) को (r) brackets से चुनना / Choosing (b) from (r) brackets. There are \(^{n}C_r\) ways to choose (b) from (r) of the (n) brackets. In exams treat a binomial coefficient as selection.
Step 3
Exam Tip
(n) brackets में से (r) brackets में (b) चुनने के \(^{n}C_r\) तरीके हैं। परीक्षा में binomial coefficient को selection मानें।
One group has (r) chosen objects and the other has (n-r) remaining objects. In exams check that group sizes add to (n).
Step 2
Why this answer is correct
The correct answer is C. (r) और (n-r). One group has (r) chosen objects and the other has (n-r) remaining objects. In exams check that group sizes add to (n).
Step 3
Exam Tip
एक समूह में (r) चुनी वस्तुएँ और दूसरे में (n-r) बाकी वस्तुएँ होती हैं। परीक्षा में group sizes का योग (n) जाँचें।
Once the committee of (5) is chosen the remaining (7) students are fixed. In exams use only one combination when the complement group is fixed automatically.
Step 2
Why this answer is correct
The correct answer is A. \(^{12}C_5\). Once the committee of (5) is chosen the remaining (7) students are fixed. In exams use only one combination when the complement group is fixed automatically.
Step 3
Exam Tip
(5) की समिति चुनते ही बाकी (7) विद्यार्थी तय हो जाते हैं। परीक्षा में complement group खुद तय हो तो केवल एक combination लगाएँ।
First choose the books and then assign display order. In exams multiply by (r!) when order follows selection.
Step 2
Why this answer is correct
The correct answer is B. \(^{5}C_2\times2!\). First choose the books and then assign display order. In exams multiply by (r!) when order follows selection.
Step 3
Exam Tip
पहले किताबें चुनें फिर display order दें। परीक्षा में selection के बाद order हो तो (r!) से गुणा करें।
There are no different posts among the representatives so order is not counted. In exams use combination for only selection.
Step 2
Why this answer is correct
The correct answer is B. \(^{7}C_3\). There are no different posts among the representatives so order is not counted. In exams use combination for only selection.
Step 3
Exam Tip
प्रतिनिधियों में कोई अलग पद नहीं है इसलिए क्रम नहीं गिना जाएगा। परीक्षा में only selection हो तो combination लगाएँ।
A. (r)वें स्थान के विकल्प/Choices for the (r)th position
Step 1
Concept
After filling the first (r-1) positions (n-r+1) choices remain for the (r)th position. In exams connect recurrence with the next-place idea.
Step 2
Why this answer is correct
The correct answer is A. (r)वें स्थान के विकल्प / Choices for the (r)th position. After filling the first (r-1) positions (n-r+1) choices remain for the (r)th position. In exams connect recurrence with the next-place idea.
Step 3
Exam Tip
पहले (r-1) स्थान भरने के बाद (r)वें स्थान पर (n-r+1) विकल्प बचते हैं। परीक्षा में recurrence को next-place idea से जोड़ें।
Dividing the factorial formulas gives \(\frac{n-r}{r+1}\). In exams ratios make adjacent combinations faster.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{n-r}{r+1}\). Dividing the factorial formulas gives \(\frac{n-r}{r+1}\). In exams ratios make adjacent combinations faster.
Step 3
Exam Tip
Factorial formula को divide करने पर \(\frac{n-r}{r+1}\) मिलता है। परीक्षा में adjacent combinations के लिए ratio से तेजी आती है।
\(^{n}C_n=^{n}C_0\) and both have value (1). In exams remember both all selected and none selected boundary cases.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}C_0\). \(^{n}C_n=^{n}C_0\) and both have value (1). In exams remember both all selected and none selected boundary cases.
Step 3
Exam Tip
\(^{n}C_n=^{n}C_0\) और दोनों का मान (1) है। परीक्षा में all selected और none selected दोनों boundary cases याद रखें।
If the special object is selected the remaining (r-1) objects are chosen from (n-1). In exams the lower index decreases by (1) in the included case.
Step 2
Why this answer is correct
The correct answer is B. \(^{n-1}C_{r-1}\). If the special object is selected the remaining (r-1) objects are chosen from (n-1). In exams the lower index decreases by (1) in the included case.
Step 3
Exam Tip
खास वस्तु चुन ली गई तो बाकी (r-1) वस्तुएँ (n-1) में से चुननी हैं। परीक्षा में included case में lower index (1) घटता है।
After removing the special object all (r) objects are chosen from (n-1). In exams the lower index remains the same in the excluded case.
Step 2
Why this answer is correct
The correct answer is A. \(^{n-1}C_r\). After removing the special object all (r) objects are chosen from (n-1). In exams the lower index remains the same in the excluded case.
Step 3
Exam Tip
खास वस्तु हटाने के बाद (n-1) वस्तुओं में से पूरे (r) चुनने हैं। परीक्षा में excluded case में lower index वही रहता है।
Including or excluding one fixed object gives \(^{9}C_3\) and \(^{9}C_4\). In exams the upper index decreases by (1) in a Pascal split.
Step 2
Why this answer is correct
The correct answer is A. \(^{9}C_4+^{9}C_3\). Including or excluding one fixed object gives \(^{9}C_3\) and \(^{9}C_4\). In exams the upper index decreases by (1) in a Pascal split.
Step 3
Exam Tip
एक fixed object को include या exclude करने से \(^{9}C_3\) और \(^{9}C_4\) मिलते हैं। परीक्षा में Pascal split में upper index (1) घटता है।
The three red balls are identical so their (3!) internal orders are not different. In exams divide by the factorial of the repeated group only.
Step 2
Why this answer is correct
The correct answer is A. (3!). The three red balls are identical so their (3!) internal orders are not different. In exams divide by the factorial of the repeated group only.
Step 3
Exam Tip
तीन लाल balls समान हैं इसलिए उनके (3!) अंदरूनी क्रम अलग नहीं हैं। परीक्षा में केवल repeated group के factorial से भाग दें।
To remove internal orders of two repeated groups divide by (2!3!). In exams put the factorial of each repeated letter in the denominator.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{8!}{2!3!}\). To remove internal orders of two repeated groups divide by (2!3!). In exams put the factorial of each repeated letter in the denominator.
Step 3
Exam Tip
दो repeated groups के internal orders हटाने के लिए (2!3!) से भाग देते हैं। परीक्षा में हर repeated letter का factorial denominator में रखें।
A. क्योंकि एक व्यक्ति को fixed मानकर rotations हटाते हैं/Because one person is fixed to remove rotations
Step 1
Concept
A mere rotation in a circle does not make a new seating. In exams use the one fixed method in circular arrangement.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि एक व्यक्ति को fixed मानकर rotations हटाते हैं / Because one person is fixed to remove rotations. A mere rotation in a circle does not make a new seating. In exams use the one fixed method in circular arrangement.
Step 3
Exam Tip
वृत्त में केवल घूमना नई seating नहीं बनाता। परीक्षा में circular arrangement में one fixed method लगाएँ।
First removing rotations gives ((n-1)!) and then same reflection makes us divide by (2). In exams check both rotation and reflection in necklaces.
Step 2
Why this answer is correct
The correct answer is C. (\frac{(n-1)!}{2}). First removing rotations gives ((n-1)!) and then same reflection makes us divide by (2). In exams check both rotation and reflection in necklaces.
Step 3
Exam Tip
पहले rotations हटाकर ((n-1)!) मिलता है फिर reflection same होने से (2) से भाग देते हैं। परीक्षा में necklace में rotation और reflection दोनों देखें।
Because (r!) is (1) only for (r=0) and (r=1). In exams check \(^{n}P_r=^{n}C_r r!\) for equality.
Step 2
Why this answer is correct
The correct answer is A. जब (r=0) या (r=1) हो / When (r=0) or (r=1). Because (r!) is (1) only for (r=0) and (r=1). In exams check \(^{n}P_r=^{n}C_r r!\) for equality.
Step 3
Exam Tip
क्योंकि (r!) केवल (r=0) और (r=1) पर (1) होता है। परीक्षा में equality के लिए relation \(^{n}P_r=^{n}C_r r!\) देखें।
A. क्योंकि combination में (r!) orders को एक ही selection माना जाता है/Because in combination (r!) orders are treated as one selection
Step 1
Concept
Permutation counts many orders of the same group. In exams keep the difference between (P) and (C) clear when (r>1).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि combination में (r!) orders को एक ही selection माना जाता है / Because in combination (r!) orders are treated as one selection. Permutation counts many orders of the same group. In exams keep the difference between (P) and (C) clear when (r>1).
Step 3
Exam Tip
Permutation एक ही group के कई orders गिनता है। परीक्षा में (r>1) पर (P) और (C) का अंतर साफ रखें।
A. जब extra repeated orders हटाने हों/When extra repeated orders must be removed
Step 1
Concept
When making combination from permutation the same selection is counted (r!) times. In exams use division principle when overcounting appears.
Step 2
Why this answer is correct
The correct answer is A. जब extra repeated orders हटाने हों / When extra repeated orders must be removed. When making combination from permutation the same selection is counted (r!) times. In exams use division principle when overcounting appears.
Step 3
Exam Tip
Permutation से combination बनाते समय एक ही selection (r!) बार गिना जाता है। परीक्षा में overcounting दिखे तो division principle लगाएँ।
Each shirt can pair with (4) pants so there are \(3\times4\) choices. In exams multiply when independent choices are needed together.
Step 2
Why this answer is correct
The correct answer is B. गुणन सिद्धांत / Multiplication principle. Each shirt can pair with (4) pants so there are \(3\times4\) choices. In exams multiply when independent choices are needed together.
Step 3
Exam Tip
हर shirt के साथ (4) pants जुड़ सकते हैं इसलिए \(3\times4\) choices हैं। परीक्षा में independent choices साथ चाहिए तो multiply करें।
This is an or situation so there are (5+6) choices. In exams look for the addition principle when the word or appears.
Step 2
Why this answer is correct
The correct answer is B. जोड़ सिद्धांत / Addition principle. This is an or situation so there are (5+6) choices. In exams look for the addition principle when the word or appears.
Step 3
Exam Tip
यह या वाली स्थिति है इसलिए (5+6) choices मिलते हैं। परीक्षा में or शब्द पर addition principle देखें।
In a committee the order of members does not change the committee. In exams use combination for committee or group.
Step 2
Why this answer is correct
The correct answer is C. समिति / Committee. In a committee the order of members does not change the committee. In exams use combination for committee or group.
Step 3
Exam Tip
समिति में सदस्यों का क्रम नहीं बदलता। परीक्षा में committee या group पर combination लगाएँ।
B. न चुनी गई वस्तुओं की संख्या/Number of unchosen objects
Step 1
Concept
After choosing (r) objects (n-r) objects remain. In exams remember this meaning for complement counting.
Step 2
Why this answer is correct
The correct answer is B. न चुनी गई वस्तुओं की संख्या / Number of unchosen objects. After choosing (r) objects (n-r) objects remain. In exams remember this meaning for complement counting.
Step 3
Exam Tip
(r) वस्तुएँ चुनने के बाद (n-r) वस्तुएँ बचती हैं। परीक्षा में complement counting के लिए यह अर्थ याद रखें।
Choosing no object gives one empty selection. In exams treat the empty case as valid count (1).
Step 2
Why this answer is correct
The correct answer is A. कोई वस्तु न चुनना / Choosing no object. Choosing no object gives one empty selection. In exams treat the empty case as valid count (1).
Step 3
Exam Tip
कोई वस्तु न चुनने का एक empty selection होता है। परीक्षा में empty case को valid count (1) मानें।
B. सभी (n) objects को क्रम में जमाना/Arranging all (n) objects in order
Step 1
Concept
When (r=n) all objects are being arranged so the count is (n!). In exams treat full arrangement as factorial.
Step 2
Why this answer is correct
The correct answer is B. सभी (n) objects को क्रम में जमाना / Arranging all (n) objects in order. When (r=n) all objects are being arranged so the count is (n!). In exams treat full arrangement as factorial.
Step 3
Exam Tip
(r=n) पर सभी वस्तुओं की व्यवस्था हो रही है इसलिए count (n!) है। परीक्षा में full arrangement को factorial मानें।
A. chosen group और rejected group/chosen group and rejected group
Step 1
Concept
Each selection has a unique rejected group. In exams understand the identity by forming a complementary pair.
Step 2
Why this answer is correct
The correct answer is A. chosen group और rejected group / chosen group and rejected group. Each selection has a unique rejected group. In exams understand the identity by forming a complementary pair.
Step 3
Exam Tip
हर selection का एक unique rejected group होता है। परीक्षा में complementary pair बनाकर identity समझें।
A. क्योंकि शुरुआत में (n) वस्तुओं को क्रम में जमाकर गिना जा सकता है/Because initially (n) objects can be counted by arranging in order
Step 1
Concept
(n!) creates a larger ordered count which is corrected by (r!(n-r)!). In exams understand the overcounting correction.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि शुरुआत में (n) वस्तुओं को क्रम में जमाकर गिना जा सकता है / Because initially (n) objects can be counted by arranging in order. (n!) creates a larger ordered count which is corrected by (r!(n-r)!). In exams understand the overcounting correction.
Step 3
Exam Tip
(n!) से ordered arrangements की बड़ी गिनती बनती है जिसे (r!(n-r)!) से सही किया जाता है। परीक्षा में overcounting correction समझें।
In a directed arrow changing start and end changes the arrow. In exams use ordered pair or permutation when direction exists.
Step 2
Why this answer is correct
The correct answer is B. \(^{5}P_2\). In a directed arrow changing start and end changes the arrow. In exams use ordered pair or permutation when direction exists.
Step 3
Exam Tip
Directed arrow में start और end बदलने से arrow बदल जाता है। परीक्षा में direction हो तो ordered pair यानी permutation लगाएँ।
