Bank Exams प्रश्न और उत्तर का अभ्यास करें

प्र: In a plane 8 points are colliner  out of 12 points, then the number of triangles we get with those 12 points is 1036 0

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    20
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  • 2
    160
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  • 3
    164
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  • 4
    220
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उत्तर : 3. "164"
व्याख्या :

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.   If all these 8 points are different we get 8C3 triangles as they are collinear.   In 12C3 triangles, we do not get 8C3 triangles   Therefore, The number of triangles we get = 12C3-8C3 = 164

प्र: If A1, A2, A3, A4, ..... A10 are speakers for a meeting and A1 always speaks after, A2 then the number of ways they can speak in the meeting is 1522 0

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    9!
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  • 2
    9!/2
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  • 3
    10!
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  • 4
    10!/2
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उत्तर : 4. "10!/2"
व्याख्या :

Answer: D) 10!/2 Explanation: As A1 speaks always after A2, they can speak only in  1st  to 9th places and    A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place    A2 can speak in 9 places the remaining     A3, A4, A5,...A10  has no restriction. So, they can speak in 9.8! ways. i.e   when A2 speaks in the first place, the number of ways they can speak is 9.8!.   When A2 speaks in second place, the number of ways they can speak is  8.8!.   When A2 speaks in third place, the number of ways they can speak is  7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!       Therefore,Total Number of ways they can  speak = (9+8+7+6+5+4+3+2+1) 8! = 92(9+1)8! = 10!/2

प्र: The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is 1280 0

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    7! x 7!
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  • 2
    7! x 6!
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  • 3
    6! x 6!
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  • 4
    7! x 5!
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उत्तर : 2. "7! x 6!"
व्याख्या :

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 7P6ways. 7 teachers can sit in (7-1)! ways.    Required no.of ways is = 7P6.6! = 7!.6!

प्र: The number of ways that 8 beads of different colours be strung as a necklace is 2193 0

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    2520
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  • 2
    2880
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    4320
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  • 4
    5040
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उत्तर : 1. "2520"
व्याख्या :

Answer: A) 2520 Explanation: The number of ways of arranging n beads in a necklace is (n-1)!2=(8-1)!2=7!2 = 2520  (since n = 8)

प्र: In how many ways the letters of the word 'DESIGN' can be arranged so that no consonant appears at either of the two ends? 1086 0

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    240
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  • 2
    72
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  • 3
    48
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  • 4
    36
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उत्तर : 3. "48"
व्याख्या :

Answer: C) 48 Explanation: DESIGN = 6 letters   No consonants appear at either of the two ends.  = 2 x 4P4 =  2 x 4 x 3 x 2 x 1=  48

प्र: In How many ways can the letters of the word 'CAPITAL' be arranged in such a way that all the vowels always come together? 1036 0

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    360
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  • 2
    720
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  • 3
    120
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  • 4
    840
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उत्तर : 1. "360"
व्याख्या :

Answer: A) 360 Explanation: CAPITAL = 7   Vowels = 3 (A, I, A)   Consonants = (C, P, T, L)   5 letters which can be arranged in  5P5=5!   Vowels A,I = 3!2!   No.of arrangements = 5! x 3!2!=360

प्र: Using numbers from 0 to 9 the number of 5 digit telephone numbers that can be formed is 1254 0

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    1,00,000
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  • 2
    59,049
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    3439
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  • 4
    6561
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उत्तर : 3. "3439"
व्याख्या :

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.   This can be done in 105ways, but '0' is there   So, the numbers starting with '0' are to be excluded is 94 numbers.    Total 5 digit telephone numbers = 105- 94 = 3439

प्र: Count the number of triangles and squares in the given figure. 2825 0

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    36 triangles, 7 Squares
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  • 2
    38 triangles, 9 Squares
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  • 3
    40 triangles, 7 Squares
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  • 4
    42 triangles, 9 Squares
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उत्तर : 3. "40 triangles, 7 Squares"
व्याख्या :

Answer: C) 40 triangles, 7 Squares Explanation: The figure may be labelled as shown      Triangles :   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   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.   The triangles composed of four components each are BHI, GJK, ILD, AGJ, HIL and JKE i.e 6 in number.   Total number of triangles in the figure = 18 + 16 + 6 =40.   Squares :   The Squares composed of two components each are MGNH, NIOJ, and OKPL i.e 3 in number   The Squares composed of four components each are BGHA, GIJH, IKJL and KDEL i.e 4 in number   Total number of squares in the figure = 3 + 4 =7

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