Bank Exams рдкреНрд░рд╢реНрди рдФрд░ рдЙрддреНрддрд░ рдХрд╛ рдЕрднреНрдпрд╛рд╕ рдХрд░реЗрдВ
рдкреНрд░: In a plane 8 points are colliner ┬аout of 12 points, then the number of triangles we get with those 12 points is 1192 05b5cc7b4e4d2b419777508ab
5b5cc7b4e4d2b419777508ab- 120false
- 2160false
- 3164true
- 4220false
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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- SingleChoice
рдЙрддреНрддрд░ : 3. "164"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: C) 164 Explanation: For a triangle, we need 3 non-collinear points. So with 12 points (when all the 12 are such that any three non-collinear is12C3. But among them 8 points are collinear. ┬а If all these 8 points are different we get┬а8C3┬аtriangles as they are collinear. ┬а In┬а12C3┬аtriangles, we do not get┬а8C3┬аtriangles ┬а Therefore, The number of triangles we get = 12C3-8C3┬а= 164
рдкреНрд░: If A1, A2, A3, A4, ..... A10┬аare speakers for a meeting and A1 always speaks after, A2 then the number of ways they can speak in the meeting is 1660 05b5cc7b4e4d2b419777508a6
5b5cc7b4e4d2b419777508a6- 19!false
- 29!/2false
- 310!false
- 410!/2true
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 4. "10!/2"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: D) 10!/2 Explanation: As A1 speaks always after A2, they can speak only in ┬а1st ┬аto 9th┬аplaces and┬а ┬а A2┬аcan speak in 2nd to 10 the places only when┬аA1┬аspeaks in 1st place┬а ┬а A2┬аcan speak in 9 places the remaining┬а ┬а ┬аA3, A4, A5,...A10┬а has no restriction. So, they can speak in 9.8! ways. i.e ┬а when A2 speaks in the┬аfirst place, the number of ways they can speak is 9.8!. ┬а When┬аA2┬аspeaks in second place, the number of ways they can speak is ┬а8.8!. ┬а When┬аA2┬аspeaks in third place, the number of ways they can speak is ┬а7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8! ┬а ┬а ┬а Therefore,Total Number of ways they can ┬аspeak = (9+8+7+6+5+4+3+2+1) 8! =┬а92(9+1)8!┬а= 10!/2
рдкреНрд░: The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is 1481 05b5cc7b4e4d2b419777508a1
5b5cc7b4e4d2b419777508a1- 17! x 7!false
- 27! x 6!true
- 36! x 6!false
- 47! x 5!false
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 2. "7! x 6!"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: B) 7! x 6! Explanation: The students should sit in between two teachers. There are 7 gaps in between teachers when they sit in a roundtable. This can be done in┬а7P6ways. 7 teachers can sit in (7-1)! ways. ┬а ┬аRequired no.of ways is┬а=┬а7P6.6!┬а=┬а7!.6!
рдкреНрд░: The number of ways that 8 beads of different colours be strung as a necklace is 2403 05b5cc7b4e4d2b4197775089c
5b5cc7b4e4d2b4197775089c- 12520true
- 22880false
- 34320false
- 45040false
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 1. "2520"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: A) 2520 Explanation: The number of ways of arranging n beads in a necklace is┬а(n-1)!2=(8-1)!2=7!2┬а= 2520┬а (since n = 8)
рдкреНрд░: In how many ways the letters of the word 'DESIGN' can be arranged so that no consonant appears at either of the two ends? 1276 05b5cc7b4e4d2b41977750892
5b5cc7b4e4d2b41977750892- 1240false
- 272false
- 348true
- 436false
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 3. "48"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: C) 48 Explanation: DESIGN = 6 letters ┬а No consonants appear┬аat either of the two ends.┬а =┬а2┬аx┬а4P4┬а= ┬а2 x 4 x 3 x 2 x 1= ┬а48
рдкреНрд░: In How many ways can the letters of the word 'CAPITAL' be arranged in such a way that all the vowels always come together? 1270 05b5cc7b4e4d2b41977750897
5b5cc7b4e4d2b41977750897- 1360true
- 2720false
- 3120false
- 4840false
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 1. "360"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: A) 360 Explanation: CAPITAL = 7 ┬а Vowels = 3 (A, I, A) ┬а Consonants = (C, P, T, L) ┬а 5 letters which can be arranged in┬а┬а5P5=5! ┬а Vowels A,I =┬а3!2! ┬а No.of arrangements = 5! x┬а3!2!=360
рдкреНрд░: Using numbers from 0 to 9 the number of 5 digit telephone numbers that can be formed is 1452 05b5cc7b4e4d2b4197775088d
5b5cc7b4e4d2b4197775088d- 11,00,000false
- 259,049false
- 33439true
- 46561false
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 3. "3439"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: C) 3439 Explanation: The numbers 0,1,2,3,4,5,6,7,8,9 are 10 in number while preparing telephone numbers any number can be used any number of times. ┬а This can be done in┬а105ways, but '0' is there ┬а So, the numbers starting with '0' are to be excluded is┬а94┬аnumbers. ┬а ┬аTotal 5 digit telephone numbers =┬а105-┬а94 = 3439
рдкреНрд░: Count the number of triangles and squares in the given figure. 3029 05b5cc7b0e4d2b41977750826
5b5cc7b0e4d2b41977750826- 136 triangles, 7 Squaresfalse
- 238 triangles, 9 Squaresfalse
- 340 triangles, 7 Squarestrue
- 442 triangles, 9 Squaresfalse
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 3. "40 triangles, 7 Squares"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: C) 40 triangles, 7 Squares Explanation: The figure may be labelled as shown┬а ┬а ┬а Triangles : ┬а The Simplest triangles are BGM, GHM, HAM, ABM, GIN, IJN, JHN, HGN, IKO, KLO, LJO, JIO, KDP, DEP, ELP, LKP, BCD and AFE i.e 18 in number ┬а The triangles composed of two components each are ABG, BGH, GHA, HAB, HGI, GIJ, IJH, JHG, JIK, IKL, KLJ,LJI, LKD, KDE, DEL and ELK i.e 16 in number. ┬а The triangles composed of four components each are BHI, GJK, ILD, AGJ, HIL and JKE i.e 6 in number. ┬а Total number of triangles in the figure = 18 + 16 + 6 =40. ┬а Squares : ┬а The Squares composed of two components each are MGNH, NIOJ, and OKPL i.e 3 in number ┬а The Squares composed of four components each are BGHA, GIJH, IKJL and KDEL i.e 4 in number ┬а Total number of squares in the figure = 3 + 4 =7

