Treat the (5) vowels as one block to get \(4!\cdot5!=2880\) arrangements. In exams, use the block method for together conditions.
Step 2
Why this answer is correct
The correct answer is B. (2880). Treat the (5) vowels as one block to get \(4!\cdot5!=2880\) arrangements. In exams, use the block method for together conditions.
Step 3
Exam Tip
(5) स्वरों को एक ब्लॉक मानकर \(4!\cdot5!=2880\) व्यवस्थाएं मिलती हैं। परीक्षा में साथ रहने की शर्त में ब्लॉक विधि लगाएं।
First arrange the non-(R) letters in \(\frac{5!}{2!}=60\) ways and place two (R)'s in \(\binom{6}{2}\) gaps. The total is \(60\cdot15=900\).
Step 2
Why this answer is correct
The correct answer is A. (900). First arrange the non-(R) letters in \(\frac{5!}{2!}=60\) ways and place two (R)'s in \(\binom{6}{2}\) gaps. The total is \(60\cdot15=900\).
Step 3
Exam Tip
पहले बिना (R) के अक्षरों की \(\frac{5!}{2!}=60\) व्यवस्थाएं बनती हैं और (6) gaps में दो (R) \(\binom{6}{2}\) तरीकों से रखे जाते हैं। कुल \(60\cdot15=900\) है।
The last digit is (0) or (5), and both cases must be counted separately. The total is \(^{7}P_4+6\cdot{}^{6}P_3=840+420=1260\).
Step 2
Why this answer is correct
The correct answer is B. (1260). The last digit is (0) or (5), and both cases must be counted separately. The total is \(^{7}P_4+6\cdot{}^{6}P_3=840+420=1260\).
Step 3
Exam Tip
अंतिम अंक (0) या (5) होगा और दोनों cases अलग-अलग गिनने होंगे। कुल \(^{7}P_4+6\cdot{}^{6}P_3=840+420=1260\) है।
First seat the men around the circle in ((5-1)!) ways and then place the women in gaps in (5!) ways. The total is \(4!\cdot5!=2880\).
Step 2
Why this answer is correct
The correct answer is A. (2880). First seat the men around the circle in ((5-1)!) ways and then place the women in gaps in (5!) ways. The total is \(4!\cdot5!=2880\).
Step 3
Exam Tip
पहले पुरुषों को गोल में ((5-1)!) तरीकों से बैठाएं और फिर महिलाओं को (5!) तरीकों से gaps में रखें। कुल \(4!\cdot5!=2880\) है।
If the unit digit is (0), there are \(6\cdot5\cdot4\) ways, and if it is (2,4,6), there are \(3\cdot5\cdot5\cdot4\) ways. The total is (120+300=420).
Step 2
Why this answer is correct
The correct answer is C. (360). If the unit digit is (0), there are \(6\cdot5\cdot4\) ways, and if it is (2,4,6), there are \(3\cdot5\cdot5\cdot4\) ways. The total is (120+300=420).
Step 3
Exam Tip
इकाई स्थान (0) हो तो \(6\cdot5\cdot4\) और इकाई स्थान (2,4,6) हो तो \(3\cdot5\cdot5\cdot4\) तरीके हैं। कुल (120+300=420) है।
Subtract together cases \(6!\cdot2!\) and both-end cases \(2!\cdot5!\) from (7!). There is no overlap because both-end positions are not together, so the result is (5040-1440-240=3360).
Step 2
Why this answer is correct
The correct answer is B. (3720). Subtract together cases \(6!\cdot2!\) and both-end cases \(2!\cdot5!\) from (7!). There is no overlap because both-end positions are not together, so the result is (5040-1440-240=3360).
Step 3
Exam Tip
कुल (7!) से साथ वाली \(6!\cdot2!\) और दोनों सिरों वाली \(2!\cdot5!\) व्यवस्थाएं घटाएं। overlap साथ नहीं हो सकता, इसलिए (5040-1440-240=3360) मिलता है।
There are (9) letters with (M,T,E) each repeated twice, so \(\frac{9!}{2!2!2!}=45360\). In exams, put each repeated letter factorial in the denominator.
Step 2
Why this answer is correct
The correct answer is A. (45360). There are (9) letters with (M,T,E) each repeated twice, so \(\frac{9!}{2!2!2!}=45360\). In exams, put each repeated letter factorial in the denominator.
Step 3
Exam Tip
(9) अक्षरों में (M,T,E) दो-दो बार हैं, इसलिए \(\frac{9!}{2!2!2!}=45360\)। परीक्षा में हर repeated letter का factorial हर में रखें।
Choose the position of the particular flag in (5) ways and fill the other (4) positions from (8) flags in \(^{8}P_4\) ways. The total is \(5\cdot{}^{8}P_4=8400\).
Step 2
Why this answer is correct
The correct answer is B. (6720). Choose the position of the particular flag in (5) ways and fill the other (4) positions from (8) flags in \(^{8}P_4\) ways. The total is \(5\cdot{}^{8}P_4=8400\).
Step 3
Exam Tip
विशेष झंडे का स्थान (5) तरीकों से चुनें और बाकी (4) स्थान (8) झंडों से \(^{8}P_4\) तरीकों से भरें। कुल \(5\cdot{}^{8}P_4=8400\) है।
(6) अलग-अलग गणित और (4) अलग-अलग विज्ञान की पुस्तकों को पंक्ति में ऐसे कितने तरीकों से रखें कि विज्ञान की सभी पुस्तकें साथ हों लेकिन गणित की पुस्तकें जरूरी नहीं?
First choose (5) people in \(\binom{8}{5}\) ways and then seat them around a circle in ((5-1)!) ways. The total is \(\binom{8}{5}\cdot4!=1344\).
Step 2
Why this answer is correct
The correct answer is A. (6720). First choose (5) people in \(\binom{8}{5}\) ways and then seat them around a circle in ((5-1)!) ways. The total is \(\binom{8}{5}\cdot4!=1344\).
Step 3
Exam Tip
पहले (5) लोगों को \(\binom{8}{5}\) तरीकों से चुनें और फिर गोल में ((5-1)!) तरीकों से बैठाएं। कुल \(\binom{8}{5}\cdot4!=1344\) है।
This is a derangement of (6) objects, whose number is (!6=265). In exams, when none is in the correct place, think derangement.
Step 2
Why this answer is correct
The correct answer is A. (265). This is a derangement of (6) objects, whose number is (!6=265). In exams, when none is in the correct place, think derangement.
Step 3
Exam Tip
यह (6) वस्तुओं का derangement है जिसकी संख्या (!6=265) होती है। परीक्षा में कोई भी सही स्थान पर न हो तो derangement सोचें।
Total valid (5)-digit numbers are \(8\cdot{}^{8}P_4\), and those without (0) are \(^{8}P_5\). The difference is (13440-6720=6720).
Step 2
Why this answer is correct
The correct answer is A. (6720). Total valid (5)-digit numbers are \(8\cdot{}^{8}P_4\), and those without (0) are \(^{8}P_5\). The difference is (13440-6720=6720).
Step 3
Exam Tip
कुल वैध (5)-अंकीय संख्याएं \(8\cdot{}^{8}P_4\) हैं और बिना (0) वाली \(^{8}P_5\) हैं। अंतर (13440-6720=6720) है।
Treat the two students as a block, count \(6!\cdot2!\) together arrangements and subtract cases where the third is at an end. The answer is (1200).
Step 2
Why this answer is correct
The correct answer is B. (1200). Treat the two students as a block, count \(6!\cdot2!\) together arrangements and subtract cases where the third is at an end. The answer is (1200).
Step 3
Exam Tip
दो छात्रों को ब्लॉक मानकर पहले साथ वाली \(6!\cdot2!\) व्यवस्थाएं लें और तीसरे के end cases घटाएं। उत्तर (1200) है।
Treat the two beads as one block, so the necklace count for (7) units is (\frac{(7-1)!}{2}), with (2!) internal ways. The total is (720).
Step 2
Why this answer is correct
The correct answer is A. (720). Treat the two beads as one block, so the necklace count for (7) units is (\frac{(7-1)!}{2}), with (2!) internal ways. The total is (720).
Step 3
Exam Tip
दो मोतियों को एक ब्लॉक मानें, तो (7) इकाइयों की necklace संख्या (\frac{(7-1)!}{2}) है और ब्लॉक के अंदर (2!) तरीके हैं। कुल (720) है।
The thousands place has (4) choices from (4,5,6,7), and the remaining places have \(8\cdot7\cdot6\) ways. The total is (1344).
Step 2
Why this answer is correct
The correct answer is B. (1680). The thousands place has (4) choices from (4,5,6,7), and the remaining places have \(8\cdot7\cdot6\) ways. The total is (1344).
Step 3
Exam Tip
हजार स्थान पर (4,5,6,7) में से (4) विकल्प हैं और बाकी \(8\cdot7\cdot6\) तरीके हैं। कुल (1344) है।
There are (11) letters with (E) three times, (N) three times, (G) twice and (I) twice. The number is \(\frac{11!}{3!3!2!2!}=277200\).
Step 2
Why this answer is correct
The correct answer is A. (277200). There are (11) letters with (E) three times, (N) three times, (G) twice and (I) twice. The number is \(\frac{11!}{3!3!2!2!}=277200\).
Step 3
Exam Tip
(11) अक्षरों में (E) तीन, (N) तीन, (G) दो और (I) दो बार हैं। संख्या \(\frac{11!}{3!3!2!2!}=277200\) है।
Arrange the men first in (6!) ways and place (4) women in (7) gaps in \(^{7}P_4\) ways. The total is \(6!\cdot{}^{7}P_4=604800\).
Step 2
Why this answer is correct
The correct answer is A. (604800). Arrange the men first in (6!) ways and place (4) women in (7) gaps in \(^{7}P_4\) ways. The total is \(6!\cdot{}^{7}P_4=604800\).
Step 3
Exam Tip
पहले पुरुषों को (6!) तरीकों से बैठाएं और (7) gaps में (4) महिलाएं \(^{7}P_4\) तरीकों से रखें। कुल \(6!\cdot{}^{7}P_4=604800\) है।
The first digit can be (3,4,5,6,7), and the remaining (4) places are filled in \(^{7}P_4\) ways. The total is \(5\cdot840=4200\).
Step 2
Why this answer is correct
The correct answer is A. (4320). The first digit can be (3,4,5,6,7), and the remaining (4) places are filled in \(^{7}P_4\) ways. The total is \(5\cdot840=4200\).
Step 3
Exam Tip
पहला अंक (3,4,5,6,7) हो सकता है और बाकी (4) स्थान \(^{7}P_4\) तरीकों से भरेंगे। कुल \(5\cdot840=4200\) है।
The particular colour has (2) possible positions, and the remaining (3) positions are filled from (6) colours in \(^{6}P_3\) ways. The total is \(2\cdot120=240\).
Step 2
Why this answer is correct
The correct answer is B. (240). The particular colour has (2) possible positions, and the remaining (3) positions are filled from (6) colours in \(^{6}P_3\) ways. The total is \(2\cdot120=240\).
Step 3
Exam Tip
विशेष रंग के लिए (2) स्थान हैं और बाकी (3) स्थान (6) रंगों से \(^{6}P_3\) तरीकों से भरेंगे। कुल \(2\cdot120=240\) है।
Choose the other (2) digits from (6) digits in \(\binom{6}{2}\) ways and arrange the (4) digits in (4!) ways. The total is \(15\cdot24=360\).
Step 2
Why this answer is correct
The correct answer is A. (720). Choose the other (2) digits from (6) digits in \(\binom{6}{2}\) ways and arrange the (4) digits in (4!) ways. The total is \(15\cdot24=360\).
Step 3
Exam Tip
बाकी (2) अंक (6) अंकों में से \(\binom{6}{2}\) तरीकों से चुनें और (4) अंकों को (4!) तरीकों से सजाएं। कुल \(15\cdot24=360\) है।
The particular student has (3) inner positions, and the remaining (4) places are filled from (8) students in \(^{8}P_4\) ways. The total is \(3\cdot1680=5040\).
Step 2
Why this answer is correct
The correct answer is B. (6720). The particular student has (3) inner positions, and the remaining (4) places are filled from (8) students in \(^{8}P_4\) ways. The total is \(3\cdot1680=5040\).
Step 3
Exam Tip
विशेष छात्र के लिए (3) अंदरूनी स्थान हैं और बाकी (4) स्थान (8) छात्रों से \(^{8}P_4\) तरीकों से भरेंगे। कुल \(3\cdot1680=5040\) है।
Treat each couple as one block, giving (5!) block arrangements and (2!) ways inside each block. The total is (5!\cdot(2!)5=3840).
Step 2
Why this answer is correct
The correct answer is B. (3840). Treat each couple as one block, giving (5!) block arrangements and (2!) ways inside each block. The total is (5!\cdot(2!)5=3840).
Step 3
Exam Tip
हर जोड़े को एक ब्लॉक मानें, तो (5!) ब्लॉक व्यवस्थाएं और प्रत्येक ब्लॉक में (2!) तरीके हैं। कुल (5!\cdot(2!)5=3840) है।
The two particular objects must have positions differing by (4), giving (4) position-pairs and (2!) orders. The remaining objects arrange in (6!) ways, so the total is \(4\cdot2!\cdot6!=5760\).
Step 2
Why this answer is correct
The correct answer is A. (2880). The two particular objects must have positions differing by (4), giving (4) position-pairs and (2!) orders. The remaining objects arrange in (6!) ways, so the total is \(4\cdot2!\cdot6!=5760\).
Step 3
Exam Tip
दो विशेष वस्तुओं के स्थानों में (4) का अंतर होना चाहिए, ऐसे (4) position-pairs हैं और order (2!) है। बाकी (6!) तरीकों से सजेंगे, कुल \(4\cdot2!\cdot6!=5760\) है।
There are (4) choices (1,3,5,7) for the last place, and the first place cannot be (0). The total is \(4\cdot7\cdot{}^{7}P_3=5880\).
Step 2
Why this answer is correct
The correct answer is A. (6720). There are (4) choices (1,3,5,7) for the last place, and the first place cannot be (0). The total is \(4\cdot7\cdot{}^{7}P_3=5880\).
Step 3
Exam Tip
अंतिम स्थान पर (1,3,5,7) के (4) विकल्प हैं और पहला स्थान (0) नहीं हो सकता। कुल \(4\cdot7\cdot{}^{7}P_3=5880\) है।
Treat the three particular people as one block, giving (5) units around a circle with ((5-1)!) arrangements and (3!) internal ways. The total is \(24\cdot6=144\).
Step 2
Why this answer is correct
The correct answer is A. (720). Treat the three particular people as one block, giving (5) units around a circle with ((5-1)!) arrangements and (3!) internal ways. The total is \(24\cdot6=144\).
Step 3
Exam Tip
तीन विशेष लोगों को एक ब्लॉक मानें, तो (5) इकाइयों की गोल व्यवस्था ((5-1)!) है और अंदर (3!) तरीके हैं। कुल \(24\cdot6=144\) है।
Subtract the cases where the particular student is secretary, \(^{9}P_3\), from total \(^{10}P_4\). The answer is (5040-504=4536).
Step 2
Why this answer is correct
The correct answer is A. (4536). Subtract the cases where the particular student is secretary, \(^{9}P_3\), from total \(^{10}P_4\). The answer is (5040-504=4536).
Step 3
Exam Tip
कुल \(^{10}P_4\) में से विशेष छात्र के सचिव बनने वाली \(^{9}P_3\) व्यवस्थाएं घटाएं। उत्तर (5040-504=4536) है।
The first digit is fixed as (3), and the last place has (4) even choices. The middle (3) places are filled in \(7\cdot6\cdot5\) ways, giving (840).
Step 2
Why this answer is correct
The correct answer is C. (360). The first digit is fixed as (3), and the last place has (4) even choices. The middle (3) places are filled in \(7\cdot6\cdot5\) ways, giving (840).
Step 3
Exam Tip
पहला अंक (3) निश्चित है और अंतिम स्थान के लिए (4) सम विकल्प हैं। बीच के (3) स्थान \(7\cdot6\cdot5\) तरीकों से भरेंगे, कुल (840) है।
Choose exactly one of the two particular books in \(\binom{2}{1}\) ways and the other (4) books from (6). Then arrange (5!) ways, giving \(2\cdot\binom{6}{4}\cdot5!=3600\).
Step 2
Why this answer is correct
The correct answer is A. (2880). Choose exactly one of the two particular books in \(\binom{2}{1}\) ways and the other (4) books from (6). Then arrange (5!) ways, giving \(2\cdot\binom{6}{4}\cdot5!=3600\).
Step 3
Exam Tip
दो विशेष पुस्तकों में से एक \(\binom{2}{1}\) तरीकों से और बाकी (4) पुस्तकें (6) में से चुनें। फिर (5!) से सजाएं, कुल \(2\cdot\binom{6}{4}\cdot5!=3600\) है।
The thousands place has (3) choices (1,2,3), and the remaining (3) places have \(8\cdot7\cdot6\) ways. The total is (1008).
Step 2
Why this answer is correct
The correct answer is A. (1344). The thousands place has (3) choices (1,2,3), and the remaining (3) places have \(8\cdot7\cdot6\) ways. The total is (1008).
Step 3
Exam Tip
हजार स्थान पर (1,2,3) के (3) विकल्प हैं और बाकी (3) स्थान \(8\cdot7\cdot6\) तरीकों से भरेंगे। कुल (1008) है।
Treat each couple as one block, so (4) blocks have ((4-1)!) circular arrangements and ((2!)4) internal ways. The total is (96).
Step 2
Why this answer is correct
The correct answer is A. (384). Treat each couple as one block, so (4) blocks have ((4-1)!) circular arrangements and ((2!)4) internal ways. The total is (96).
Step 3
Exam Tip
हर जोड़े को एक ब्लॉक मानें, तो (4) ब्लॉकों की गोल व्यवस्था ((4-1)!) है और अंदर ((2!)4) तरीके हैं। कुल (96) है।
The particular student has (3) non-first positions, and the other (3) positions are filled from (7) students in \(^{7}P_3\) ways. The total is \(3\cdot210=630\).
Step 2
Why this answer is correct
The correct answer is B. (1260). The particular student has (3) non-first positions, and the other (3) positions are filled from (7) students in \(^{7}P_3\) ways. The total is \(3\cdot210=630\).
Step 3
Exam Tip
विशेष छात्र के लिए अंतिम (3) positions हैं और बाकी (3) positions (7) छात्रों से \(^{7}P_3\) तरीकों से भरते हैं। कुल \(3\cdot210=630\) है।
Treat (1) and (8) as a block and choose the other (3) digits from (6). The total is \(\binom{6}{3}\cdot4!\cdot2!=960\).
Step 2
Why this answer is correct
The correct answer is B. (1800). Treat (1) and (8) as a block and choose the other (3) digits from (6). The total is \(\binom{6}{3}\cdot4!\cdot2!=960\).
Step 3
Exam Tip
(1) और (8) को ब्लॉक मानें और बाकी (3) अंक (6) में से चुनें। कुल \(\binom{6}{3}\cdot4!\cdot2!=960\) है।
Subtract cases where the three particular books are together, \(7!\cdot3!\), from total (9!). The answer is (362880-30240=332640).
Step 2
Why this answer is correct
The correct answer is A. (322560). Subtract cases where the three particular books are together, \(7!\cdot3!\), from total (9!). The answer is (362880-30240=332640).
Step 3
Exam Tip
कुल (9!) में से तीनों विशेष पुस्तकों के साथ वाले \(7!\cdot3!\) cases घटाएं। उत्तर (362880-30240=332640) है।
Treat the four (S)'s as one block, giving (8) units with (I) four times and (P) twice. The number is \(\frac{8!}{4!2!}=840\).
Step 2
Why this answer is correct
The correct answer is A. (1260). Treat the four (S)'s as one block, giving (8) units with (I) four times and (P) twice. The number is \(\frac{8!}{4!2!}=840\).
Step 3
Exam Tip
चार (S) को एक ब्लॉक मानें, तो (8) इकाइयों में (I) चार और (P) दो बार हैं। संख्या \(\frac{8!}{4!2!}=840\) है।
Total circular arrangements are (7!). If the particular object has the two given objects as neighbours, they can be placed around it in (2!) ways and the rest in (5!) ways. The answer is (5040-240=4800).
Step 2
Why this answer is correct
The correct answer is A. (4320). Total circular arrangements are (7!). If the particular object has the two given objects as neighbours, they can be placed around it in (2!) ways and the rest in (5!) ways. The answer is (5040-240=4800).
Step 3
Exam Tip
कुल गोल व्यवस्थाएं (7!) हैं। विशेष वस्तु के दोनों पड़ोसी वे दो दी गई वस्तुएं हों तो उन्हें विशेष वस्तु के पास (2!) तरीकों से रखकर बाकी (5!) तरीके मिलते हैं। उत्तर (5040-240=4800) है।
There are (5) odd choices for the first digit and (5) even choices for the last digit. The middle (4) places are filled from (8) digits in \(^{8}P_4\) ways, giving \(5\cdot5\cdot1680=42000\).
Step 2
Why this answer is correct
The correct answer is B. (67200). There are (5) odd choices for the first digit and (5) even choices for the last digit. The middle (4) places are filled from (8) digits in \(^{8}P_4\) ways, giving \(5\cdot5\cdot1680=42000\).
Step 3
Exam Tip
पहले अंक के लिए (5) विषम विकल्प और अंतिम अंक के लिए (5) सम विकल्प हैं। बीच के (4) स्थान (8) अंकों से \(^{8}P_4\) तरीकों से भरेंगे, कुल \(5\cdot5\cdot1680=42000\) है।
(8) अलग-अलग व्यक्तियों की पंक्ति व्यवस्था में पहला विशेष व्यक्ति दूसरे विशेष व्यक्ति से पहले और तीसरा विशेष व्यक्ति चौथे विशेष व्यक्ति से पहले हो, तो कितनी व्यवस्थाएं होंगी?
Out of total (8!) arrangements, each independent order condition keeps half the arrangements. Hence the answer is \(\frac{8!}{2\cdot2}=10080\).
Step 2
Why this answer is correct
The correct answer is C. (10080). Out of total (8!) arrangements, each independent order condition keeps half the arrangements. Hence the answer is \(\frac{8!}{2\cdot2}=10080\).
Step 3
Exam Tip
कुल (8!) व्यवस्थाओं में दोनों स्वतंत्र order conditions आधी-आधी व्यवस्थाएं रखती हैं। इसलिए उत्तर \(\frac{8!}{2\cdot2}=10080\) है।
